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1. What is the units digit of 7 multiplied by itself 2026 times, that is 7 to the power 2026?[5]
A it cannot be found without a calculatorB 1C 3D 7E 92. The numbers 1 to 9 are placed in a row in some order. Can every neighbouring pair add up to an odd number?[5]
A Yes, and many orders workB Yes, but only one order worksC No, there are too many odd numbersD No, nine places is an odd number of placesE Only if the row starts with an even number3. A and B are different digits. The two-digit number AB added to the two-digit number BA gives 132. What is A + B?[5]
4. How many two-digit numbers have both of their digits odd?[5]
5. What is the last digit of 3 multiplied by itself 2026 times, that is, of 3 to the power 2026?[5]
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1. E — 9 KA-0037 Last-digit behavior of products and powers
A I assumed a number this large has no reachable last digit and gave up on the cycle.B I found the cycle 7, 9, 3, 1 but used a remainder of 0 when the remainder is actually 2.C I found the cycle but counted its positions starting at zero, shifting my answer along by one.D I assumed the units digit of any power of 7 is always 7.2. A — Yes, and many orders work KA-0049 Parity of numbers (odd/even behavior)
B I found one arrangement that works and assumed a problem this fiddly could only have one answer.C I saw five odd numbers against four even ones and called that a mismatch, when a row of nine needs exactly five of one and four of the other.D I blamed the length of the row rather than checking how many odd and even numbers it has to hold.E I decided the first number settles it without checking that starting even would need five even numbers, and only four exist.3. D — 12 KA-0060 Operation puzzles and cryptarithms
A I added the digits of 132 instead of working out what A and B must be.B I found A + B correctly and then halved it, as if the question wanted one digit.C I spotted the 11 in the working and gave that instead of the sum it multiplies.E I found a pair of digits by trial and misadded them by one.4. C — 25 KA-0135 Digit-constraint puzzles
A I counted the odd numbers in one row of ten and treated that as the whole answer.B I used four odd digits instead of five, forgetting that 9 is odd.D I counted every two-digit number that is itself odd, which only fixes the last digit.E I let the first digit be any of ten values rather than only the five odd ones.5. B — 9 KA-0158 Last-digit behavior of products and powers
A I used the first entry of the cycle because 2026 is even, without working out where in the cycle it lands.C I counted the cycle starting from zero, so a remainder of 2 put me one place too far along.D I assumed a power that large must end in 1.E I multiplied 3 by 2 and used the last digit of that.