Number & Arithmetic

Half of \(1.6 \times 10^6\) equals \(0.8 \times 10^6 = 0.8 \times 10 \times 10^5 = 8 \times 10^5\).

The value of \(\dfrac{2468 \times 2468}{2468 + 2468}\) is \(\dfrac{2468 \times 2468}{2 \times 2468} = \dfrac{2468}{2}\).

Hence the correct answer is 1234.

The information on the signs pointing to Atown and the signs pointing to Cetown both tell us that the distance between the signs is (7 − 2) km = (9 − 4) km = 5 km. Therefore the distance which is written on the broken sign is (5 − 4) km = 1 km.

The sum of the row totals is the sum of all the nine numbers in the 3 × 3 square. Likewise, the sum of the column totals is the sum of these nine numbers. Therefore \(24 + 26 + 40 = 27 + 20 + x\), where \(x\) is the sum of the numbers in the third column. Therefore \(90 = 47 + x\) and hence \(x = 43\). Therefore the sum of the numbers in the third column is 43.

The values of the five options are 64, 243, 256, 125, 36 respectively.
Of these 256 is the largest, so \(4^4\) is the largest of the options.

The next year with this property is 2121 which is 101 years after 2020.

The result furthest from zero is obtained by multiplying the three numbers furthest from zero, namely -5, 4 and 6. This gives -120, which is negative and hence is the smallest possible result.

The four integers cannot all be non-positive since the total is positive. Also, they cannot all be non-negative since the smallest possible total would then be 0 + 1 + 2 + 3 = 6. So, they must have at least one negative integer, and at least one positive integer, and hence will include -1, 0 and 1. Since these have a total of 0, the fourth number must be 2. Thus the least integer is -1.

The numbers on the left use the digits M and K in the Units column, and the digits J and L in the Tens column. The numbers on the right use the digits M and K twice in the Units columns, and the digits J and L twice in the Tens column, so the total is exactly twice that of the numbers on the left. Twice 79 is 158, answer B.

The answers are 23457, 3468, 579, 1290 and 12351 respectively, so in ascending order the middle one is B.

The value of \(2 - (3 - 4) - (5 - 6 - 7) = 2 - (-1) - (-1 - 7) = 2 + 1 - (-8) = 2 + 1 + 8 = 11\).

Liam spent 50 × £1 = £50 buying the bottles. After selling 40 bottles he had £(50 + 10) = £60. Therefore he sold each bottle for £60 ÷ 40 = £1.50. Hence, after selling all 50 bottles, he then had 50 × £1.50 = £75.

The two-digit integers start with the pair of triples (101), (112). If we continue to consider the triples in pairs, then the following pairs start with the integers 13, 16, 19, etc. That is, they start with those integers that are 1 more than a multiple of 3. All of 22, 43, 46, and 88 have that form, they start the triples (222), (434), (464) and (888). Although 76 also has that form, it starts the triples (767) followed by (778), so that 777 is not a triple.

Let the missing digits be \(P\), \(Q\), \(R\). Placing the numbers in a column addition, we get:

There is nothing to carry from the sum of the tens digits, so the sum of the middle digits is \(7 + P + 3\) and this must end in 2. Hence \(P = 2\). The sum of the digits in the next column is \(5 + 2 + R + 1\) (where the 1 is carried from the middle digits). This must end in a 7, giving \(R = 9\). The first digits have sum \(1 + 2 + Q + 1\) and must end in 5 so \(Q = 1\). Therefore the missing digits are 1, 2, 9.

The value of \((7 - 6 \times (-5)) - 4 \times (-3) \div (-2) = 7 + 30 + 12 \div (-2) = 37 + (-6) = 31\).

The required sum can be written as shown below, with \(a\), \(b\) and \(c\) as the missing digits:

The sum of the digits in the units column is 16 and hence there is a carry of 1 to the tens column. Therefore, when we consider the tens column, we have \(4 + a + 2 + 1 = 2\) or \(2 + 10\). Hence \(7 + a = 2\) or 12 and, since \(a\) is a positive single-digit integer, \(a = 5\) and there is a carry of 1 to the hundreds column. Similarly, when we consider the hundreds column, we have \(2 + 1 + c + 1 = 1\) or 11 and hence \(c = 7\) and there is a carry of 1 to the thousands column. Finally, when we consider the thousands and ten thousands columns, we have \(1 + 2 + b + 1 = 10\) and hence \(b = 6\). Therefore the missing digits are 5, 6 and 7.

Let \(J\) kg be the weight of the lightest kangaroo. Then the total weight of the three kangaroos is at least \(3J\) kg. And we know they weigh 97 kg in total, so \(3J \leq 97\), so \(J \leq 32\tfrac{1}{3}\). Hence the lightest is at most 32 kg. The three could have weights 32 kg, 32 kg and 33 kg.

Each of the five friends gave a cake to the four other people. Therefore the number of cakes given away and then eaten was 4 × 5 = 20. Since this decreased the total number of cakes they had by half, the total number of cakes they had at the start was 2 × 20 = 40.

Since 2 + 0 + 1 + 9 = 12 and by inserting brackets only + or − signs are possible between the numbers, no value greater than 12 may be obtained. Also, 2 − (0 − 1 − 9) = 2 − 0 + 1 + 9 = 12 and hence a result of 12 is possible. Therefore the largest possible value that can be obtained is 12.

Note that 20 × 19 + 20 + 19 = 20 × 20 + 19 = 400 + 19 = 419.

Let the points \(O\), \(P\) etc, be labelled as \(A_1\), \(A_2\), etc, in the order in which they are drawn. Without loss of generality, we can draw the line horizontally with \(A_2\) on the right of \(A_1\). Since \(A_1\) is the midpoint of \(A_2\) and \(A_3\), we then put \(A_3\) to the left of \(A_1\) at a distance of 1 from \(A_1\). Since \(A_2\) is the midpoint of \(A_3\) and \(A_4\), we put \(A_4\) on the right of \(A_2\) at a distance of 2 from \(A_2\). Proceeding in this fashion, the odd points end up further to the left of \(A_1\), and the even points further to the right, with the distances doubling at each stage. The diagram shows the points and the distances between neighbouring points, though not to scale! The distance \(A_1A_{12}\) is \(1 + 2 + 8 + 32 + 128 + 512 = 683\).

Let the integers be \(\{2018, a_1, a_2, \ldots, a_n\}\). Their product is \(2018 \times a_1 \times a_2 \times \ldots \times a_n = 2018\) so we get \(a_1 \times a_2 \times \ldots \times a_n = 1\), meaning each \(a_i\) is either 1 or −1 (with an even number of occurrences of −1).
Their sum is \(2018 + a_1 + a_2 + \ldots + a_n = 2018\) so \(a_1 + a_2 + \ldots + a_n = 0\), meaning an equal number of occurrences of 1 and −1. Hence the list includes 2018 once, an even number of 1s and the same even number of −1s. Therefore the number of integers must be one more than a multiple of 4, so from the list of options only 2017 is possible.

Square both sides to get \(2018^2 + 2018^2 + \ldots + 2018^2 = 2018^{20}\).
Let \(n\) be the number of occurrences of \(2018^2\). Then we get \(2018^2 \times n = 2018^{20}\).
So \(n = \dfrac{2018^{20}}{2018^2} = 2018^{18}\).

The total of the sums of the three rows of the table is equal to the sum of all the digits from 1 to 9, which is 45. Similarly, the total of the sums of the three columns of the table is also equal to 45. Hence James' six answers add to 45 + 45 = 90. The sum of the five answers given is 12 + 13 + 15 + 16 + 17 = 73 and hence his sixth answer is 90 – 73 = 17.
[It is left as an exercise to find a possible arrangement of the digits 1 to 9 that actually gives these six sums.]

When the given calculations are performed on Harry's calculator, they will be equal to: A 97 × 79; B 98 × 78; C 369 × 147; D 321 × 123; E 357 × 753.
All of these, except E, are equal to the intended calculations.

Let the three missing digits from left to right be \(a\), \(b\) and \(c\). Consider the final digit ‘2’ of the answer. This is the last digit of \(3 \times b\) and hence \(b = 4\). Also, note that if \(a\) were 2 or more, then the answer would be more than 400. Therefore \(a = 1\). Hence the multiplication is 13 × 24 = 312, giving \(c = 1\). Therefore the sum of the digits scribbled out is \(a + b + c = 1 + 4 + 1 = 6\).

From the Units digit we see that \(5 + S\) ends in a 4, so \(S = 9\) (and 1 is carried). Then the Tens digit has \(4 + R + 1 = 5\) so \(R = 0\) (and nothing carried). Finally, the Hundreds digit gives \(P + Q = 6\). Then \(P + Q + R + S = 6 + 0 + 9 = 15\).

The largest possible value of \(a - b\) comes from subtracting the smallest possible value of \(b\) from the largest possible value of \(a\). Hence the largest possible value is 10 − (−5) = 10 + 5 = 15.

Let \(n\) be the middle number. Then the five numbers are \(n - 2\), \(n - 1\), \(n\), \(n + 1\), \(n + 2\) and have sum \(5n = 10^{2018}\).
Therefore \(n = \dfrac{10^{2018}}{5} = \dfrac{10 \times 10^{2017}}{5} = 2 \times 10^{2017}\).

The total face value of the £5 notes is £5 × 440 million = £2200 million = £2 200 000 000.

The positive sums of cubes less than the largest option, 36, are \(0^3 + 1^3 = 1\); \(1^3 + 2^3 = 1 + 8 = 9\) and \(2^3 + 3^3 = 8 + 27 = 35\). Hence the correct answer is 9.

Each ‘triple’ consisting of a cell and the two cells immediately below can have at most two odds (for if the bottom two are both odd, the one above is even, so they cannot be all odd). The whole diagram can be dissected into six of these (shaded) triples as shown in the top diagram, with three other (white) cells left over. These six triples have at most \(6 \times 2 = 12\) odds between them. Moreover, the three remaining white cells cannot all be odd; if we assign the values \(A\) and \(C\) to the lowest of these white cells, and \(B\) to the cell between them, then the cells above have values \(A + B\) and \(B + C\). The top white cell then contains \(A + 2B + C\), which is even when \(A\) and \(C\) are both odd. Hence the three white cells have at most two odds, giving the whole diagram at most 12 + 2 = 14 odds.
The second diagram shows one possible way of achieving this maximum of 14 odds.

Since 5 and 2 differ by 3, Tom can obtain the same integer from two different integers in the first list that also differ by 3 by adding 5 to the smaller integer and adding 2 to the larger integer. In the first list there are six pairs of integers that differ by 3, namely 1 and 4, 2 and 5, 3 and 6, 4 and 7, 5 and 8 and 6 and 9. However, the integers 4, 5 and 6 appear in two of these pairs and hence the same integer in the second list can be obtained from only three pairs of integers from the first list leaving three integers in the first list unpaired. Therefore, the smallest number of different integers Tom can obtain in the second list is six.

The three squares will be approximately a third of 770, so roughly 250. Adding up \(15^2 = 225\), \(16^2 = 256\) and \(17^2 = 289\) gives the total 770.
Alternatively, using algebra we could let \(n\) be the middle integer and then add the squares \((n - 1)^2 + n^2 + (n + 1)^2 = n^2 - 2n + 1 + n^2 + n^2 + 2n + 1 = 3n^2 + 2 = 770\).
This gives \(n^2 = 256\) and hence \(n = 16\), so the largest integer is 17.

If exactly one of \(N\) and \(N + 20\) has four digits, then the other has either 3 or 5 digits. If \(N\) has 3 digits and \(N + 20\) has 4 digits, then \(980 \leqslant N \leqslant 999\), giving 20 possibilities. If \(N\) has 4 digits and \(N + 20\) has 5 digits, then \(9980 \leqslant N \leqslant 9999\), giving 20 possibilities. Overall there are 40 possibilities for \(N\).

The total amount of the money the five girls have is (20 + 4 × 10) euros = 60 euros. Therefore, if all five girls are to have the same amount, they need to have (60 ÷ 5) euros = 12 euros each. Since each of Olivia's sisters currently has 10 euros, Olivia would need to give each of them (12 − 10) euros = 2 euros.

The sum of the three smallest positive integers is 1 + 2 + 3 = 6. Hence the only way to add three different positive integers to obtain a total of 7 is 1 + 2 + 4. Therefore the product of the three integers is 1 × 2 × 4 = 8.

Since Theo thinks it is 12:00, his watch shows 12:05. Therefore, the correct time is 12:15 since Theo's watch is 10 minutes slow. Since Leo's watch is five minutes fast, at 12:15 his watch will show 12:20. However, Leo thinks his watch is 10 minutes slow so he thinks the time is 12:30.

The sum of the numbers in the envelopes is 255. Let \(E\) be the sum of Evie's numbers. Then Alie's numbers will have a total of \(E - 31\). Hence we have \(E + (E - 31) = 255\), giving \(2E - 31 = 255\), so \(2E = 286\) and \(E = 143\). The only way to add up powers of two to get 143 is 128 + 8 + 4 + 2 + 1, so Evie took 5 envelopes.

A standard die has a total of 21 dots on its faces. The faces that are glued together have the same number of dots. Since the die in the centre of the solid has all its faces glued to other dice, the sum of the dots that are not on the surface of the solid is 2 × 21. Therefore, the number of dots on the surface of the solid is 7 × 21 − 2 × 21 = 5 × 21 = 105.

In the seven lines each of the integers from 1 to 7 is used twice and each of the integers from 8 to 14 is used once. So the sum of the numbers in the seven lines is (1 + 2 + … + 14) + (1 + 2 + … + 7) = 105 + 28 = 133. Therefore the total of the numbers in each line is 133 ÷ 7 = 19.

It is left as an exercise for the reader to show that it is possible to complete the diagram so that the total of the three numbers in each line is indeed 19.

2.017 × 2016 − 10.16 × 201.7 = 201.7 × 20.16 − 10.16 × 201.7
= 201.7(20.16 − 10.16) = 201.7 × 10 = 2017.

Any odd score must be the sum of an even (positive) number and an odd (negative number). The largest odd total will be the largest even number added to the smallest odd number, which is 6 + −1 = 5. Hence Marie cannot achieve 7. The others are achievable: 3 = 4 + −1; 4 = 2 + 2; 5 = 6 + −1; 8 = 4 + 4.

In Lucas's notation, the first zero represents the minus sign, and the number of other zeroes is the magnitude of the number, so 000 is −2 and 0000 is −3. Then 000 + 0000 is −2 + −3 = −5, which is 000000 in Lucas's notation.

Jenny subtracted 26 instead of adding 26 and obtained −14. Therefore to obtain the answer she should have obtained, she must add two lots of 26 to −14. Therefore the number she should have obtained is −14 + 2 × 26 = −14 + 52 = 38.

The calculation can be approximated as follows:
\[\frac{17 \times 0.3 \times 20.16}{999} \approx \frac{17 \times 3 \times 2}{1000} = \frac{51 \times 2}{1000} \approx \frac{100}{1000} = 0.1.\]

6116 − 2016 = 4100, so 6102 − 2016 = 4100 − 14 = 4086.

Since \(2^7 = 128\) is larger than 100, the only powers we can choose from are the first seven powers: \(2^0\) to \(2^6\), i.e. 1, 2, 4, 8, 16, 32, 64. The sum of all seven is 127. We need to eliminate one of these, and be left with a total under 100, so the only possibilities to remove would be 32 or 64. Hence there are two options:
1 + 2 + 4 + 8 + 16 + 32 = 63, and 1 + 2 + 4 + 8 + 16 + 64 = 95.

The maximum number of leaves the bush could have is 10 × 5 = 50. Each branch that has two leaves and a flower instead of five leaves reduces the number of leaves the bush has by three. Therefore the total number of leaves the bush has is of the form \(50 - 3n\) where \(n\) is the number of branches with two leaves and a flower. It is straightforward to check that none of options A to D has that form and so the answer is E.

The sum of the digits on the left-hand side of the equation is 24. Hence the equation formed by inserting + and − signs must be equivalent to 12 − 12 = 0. The smallest number of the given digits required to make 12 is three (2 + 5 + 5). Therefore the smallest number of asterisks that can be replaced by + is two and they would be placed in front of two of the three 5s.

The perimeter of the equilateral triangle is (6 + 10 + 11) cm = 27 cm. Hence the length of the sides of the equilateral triangle is 27 cm ÷ 3 = 9 cm.

Round each number in the product to one significant figure to give 2 × 500 = 1000. Hence 1000 is closest to the given product.

The units digits of \(2015^2\), \(2015^0\), \(2015^1\), \(2015^5\) are 5, 1, 5, 5, which add to 16. Thus, the units digit of the sum is 6.

The number of minutes in one week is 7 × 24 × 60, which may be written as 7 × (6 × 4) × (5 × 3 × 2 × 2) = (7 × 6 × 5 × 4 × 3 × 2) × 2. So the number of weeks in 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 minutes is 8 ÷ 2 = 4.

In one complete cycle of 4 hours, the clock is struck 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 times. So in 24 hours the clock is struck 6 × 36 = 216 times.

\((10^{640} - 1)\) is a 640-digit number consisting entirely of nines. So \(\dfrac{(10^{640} - 1)}{9}\) is a 640-digit number consisting entirely of ones.
Therefore \(\dfrac{10^{641} \times (10^{640} - 1)}{9}\) consists of 640 ones followed by 641 zeros. So \(\dfrac{10^{641} \times (10^{640} - 1)}{9} + 1\) consists of 640 ones followed by 640 zeros followed by a single one. Therefore it has 1281 digits.

In order, the values of the expressions given are: 1 − 0 = 1; 2 − 1 = 1; 9 − 8 = 1; 64 − 81 = −17; 625 − 1024 = −399.

In the rally, approximately 90 shots were hit per minute for a total of 132 minutes. As 90 × 130 = 11 700, D is the best alternative.

The digits to be used must be 9, 8, 7, 6. If any of these were to be replaced by a smaller digit, then the sum of the two two-digit numbers would be reduced. For this sum to be as large as possible, 9 and 8 must appear in the ‘tens’ column rather than the ‘units’ column. So the largest possible sum is 97 + 86 or 96 + 87. In both cases the sum is 183.

The following are the only triples of positive integers which sum to 7:

In only one of these are the three integers all different, so the required integers are 1, 2, 4 and their product is 8.

\(3 = \dfrac{2 + 2^2}{2}\); \(10 = \dfrac{4 + 4^2}{2}\); \(15 = \dfrac{5 + 5^2}{2}\); \(21 = \dfrac{6 + 6^2}{2}\). However \(\dfrac{7 + 7^2}{2} = 28\) and \(\dfrac{8 + 8^2}{2} = 36\), so 30 is not exactly halfway between a positive integer and its square.
(Note that every number which is exactly halfway between a positive integer and its square is a triangle number. Can you explain why this is so?)

The next palindromic number after 24942 is 25052, so the car travelled 110 miles in the two days.

12 ÷ 3 = 4, but this is the example given in the question. 23 ÷ 4 ≠ 5; 34 ÷ 5 ≠ 6; 45 ÷ 6 ≠ 7. However, 56 ÷ 7 = 8, so \(s\) = 8.
{Note also that 67 ÷ 8 ≠ 9.}

The diameter is (5 × 0.3 + 2.1) mm = (1.5 + 2.1) mm = 3.6 mm.

4.5 × 5.5 + 4.5 × 4.5 = 4.5(5.5 + 4.5) = 4.5 × 10 = 45.

If the difference between \(\sqrt{n}\) and 7 is less than 1, then \(6 < \sqrt{n} < 8\).
Therefore \(36 < n < 64\), so there are 27 possible values of \(n\).

The number of calories saved per day is \(\dfrac{7000}{365} \approx \dfrac{7000}{350} = 20\).

10 + 10 × 10 × (10 + 10) = 10 + 10 × 10 × 20 = 10 + 2000 = 2010.

1 + 2³ + 4 × 5 = 1 + 8 + 20 = 29.

Both £4.20 and £7.70 are multiples of 70p, so £\(C\) must also be a multiple of 70p. Of the options given, only £91 is a multiple of 70p, but it remains to check that a total cost of £91 is possible. If there are 7 children and 8 adults, then the total cost is 7 × £4.20 + 8 × £7.70 = £29.40 + £61.60 = £91.

The five options are: A 0.20088888… B 0.20080000… C 0.20080808… D 0.20080080… E 0.20082008….
So in ascending order they are B D C E A.

The difference between the largest and smallest of 9 consecutive whole numbers is 8, irrespective of the sum of the numbers. However in this case, since the average of the 9 numbers is 223, we can say they are 219, 220, …, 226, 227.

(12340 + 12.34) ÷ 1234 = 12340 ÷ 1234 + 12.34 ÷ 1234 = 10 + 0.01 = 10.01.

The date 18 years after 15 December 2005 will be 15 December 2023. Since a further 0.6 years, i.e. just over 7 months, will then elapse, the moon will next reach its highest point in the sky in 2024.

The lunch bill and tip total £28, so Gill and her friend should pay £14 each. As Gill has paid £25.50, she should now receive £25.50 − £14, that is £11.50.

The values of the expressions are, respectively, 202, 2005, 9010, 27 217 and 65 026.

In increasing order, the numbers are 4.004, 4.04, 4.044, 4.4, 4.44.

As \(P\) and \(Q\) both lie between 0 and 1, their product will be greater than 0 but smaller than \(P\) and smaller than \(Q\). Of the options available, only \(B\) satisfies these conditions. (Furthermore, its position is correct since \(P\) is approximately equal to \(\tfrac{1}{2}\), which means that the product of \(P\) and \(Q\) lies approximately half way between 0 and \(Q\).)

We note first that \(y = 5\) since that is the only non-zero digit which, when it is multiplied by 3, has itself as the units digit. So there is a carry of 1 into the tens column. We note also that \(a = 1\) or \(a = 2\) as “\(fly\)” < 1000 and therefore 3 × “\(fly\)” < 3000. We now need 3 × \(l\) + 1 to end in either 1 or 2 and the only possibility is \(l = 7\), giving \(a = 2\) with a carry of 2 into the hundreds column. As \(a = 2\), \(f\) must be at least 6. However, if \(f = 6\) then \(w = 0\) which is not allowed. Also the letters represent different digits, so \(f \neq 7\) and we can also deduce that \(f \neq 9\) since \(f = 9\) would make \(w = 9\).
Hence \(f = 8\), making \(w = 6\) and the letters represent 875 × 3 = 2625.

The sum is what we would write as 162 + 257 and this equals 419. However, in Niatirb it would be written 580.

37 373 + 61 392 = 98 765 and 45 678 + 53 087 = 98 765.

4004 − 2004 = 2000, so 4002 − 2004 = 2000 − 2 = 1998.