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1. Today is Tuesday. What day of the week will it be in 100 days?[3]
A MondayB TuesdayC WednesdayD ThursdayE Friday2. What is the remainder when 7 to the power 100 is divided by 5?[3]
3. The mean of five numbers is 12. One of the numbers is removed and the mean of the remaining four is 11. What number was removed?[4]
4. Which of these numbers can be divided exactly by both 4 and 9?[4]
5. 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 9Kangaroo Atlas · https://kangaroo-atlas.vercel.app · seed 2 · CC BY-NC-SA 4.0 · Independent project, not affiliated with Math Kangaroo in USA, NFP.
1. D — Thursday KA-0021 Remainders and modular cycles
A I found the remainder 2 but counted the two days starting from today rather than starting from tomorrow.B I divided 100 by 7 and used the quotient of 14, treating the leftover days as if there were none.C I remembered there was a remainder but used a remainder of 1 instead of working it out.E I counted 100 divided by 7 as 14 remainder 3 instead of remainder 2.2. A — 1 KA-0169 Remainders and modular cycles
B I reduced 7 to 2 correctly but then used a remainder of 1 when dividing 100 by 4.C I read the cycle backwards and landed on the entry before the right one.D I used 100 divided by 4 giving 25 and read off the 25th entry as if the cycle had length 25.E I assumed a power that large must be divisible by 5.3. D — 16 KA-0033 Means, and working backwards from a mean
A I subtracted the two means from each other and gave the difference as the removed number.B I assumed the removed number must be the new mean, since the mean dropped to 11.C I assumed removing a number equal to the old mean is what changes the mean.E I added the two totals instead of subtracting one from the other.4. D — 144 KA-0063 Divisibility rules: 2, 3, 4, 5, 9, 10
A I checked that it divides by 4 and stopped without testing the 9.B I checked the digit sum for 9 and stopped without testing the 4.C I saw an even number with a digit sum of 6 and treated 6 as good enough for 9.E I tested 9 with the digit sum and assumed any odd multiple of 9 also divides by 4.5. 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.