About Numbers
This topic collects the short cuts that make arithmetic quick. Last digits of powers, divisibility, and the standard series sums are the three that appear most often, and each has a rule that removes the need for long calculation.
What you need to understand
- The unit digit of a power depends only on the unit digit of the base, and the pattern of digits repeats.
- The cycle of unit digits is at most four long, so only the position of the power within the cycle matters.
- A number is divisible by 3 if the sum of its digits is divisible by 3, and by 9 if the digit sum is divisible by 9.
- A number is divisible by 4 if its last two digits form a multiple of 4, and by 8 if its last three digits do.
- The sum of the first n natural numbers is n(n+1)/2, of the first n even numbers n(n+1), and of the first n odd numbers n squared.
- The smallest number that has to be added to make a number divisible by d is d minus its remainder, and never d itself.
Formulas to remember
How to work through these questions
- For a unit digit question, write out the first four powers of the base and find the repeating block.
- Divide the power by the length of the cycle and use the remainder to locate the digit.
- For a divisibility question, find the remainder first and then work out how far the number falls short of the next multiple.
- For a series, decide which of the three standard sums applies before substituting anything.
- Check a series sum against a small case worked out by hand.
Mistakes that cost marks
- Dividing the power by 4 out of habit when the cycle for that base is only two or one long.
- Giving the remainder instead of the amount that still has to be added.
- Adding a whole divisor when only part of one is needed.
- Using the natural number formula for the even or the odd series.
- Assuming the unit digit of a power is the base itself.
Worked example
What is the unit digit of 7 raised to the power 45?
- The unit digits of powers of 7 are 7, 9, 3, 1 and then they repeat, so the cycle is four long.
- Divide the power by the cycle length: 45 divided by 4 leaves a remainder of 1.
- A remainder of 1 means the first entry of the cycle.
- So the unit digit is 7.
Answer: 7
Practice questions with answers
A few Numbers questions with the full solution shown, so you can see
how the method is applied before you attempt the timed set.
Question 1
Find the digit in the units place of $2^{60}$.
Answer: Option A — with explanation
The unit digit of powers of 2 repeats in a cycle, so only the position of 60 within that cycle matters.
The unit digits run through 2, 4, 8, 6, a cycle of length 4.
The power 60 sits at position 4 in that cycle, which is the digit 6.
Common mistakes
- counted one place too few round the cycle of unit digits: 5
- counted one place too far round the cycle of unit digits: 7
- counted two places too far round the cycle: 8
Question 2
What is the smallest number that must be added to 5572 to make it exactly divisible by 13?
Answer: Option B — with explanation
Dividing 5572 by 13 leaves a remainder of 8.
So 5572 is short of the next multiple by 13 - 8 = 5.
Check: 5572 + 5 = 5577, and 5577 / 13 = 429 exactly.
Common mistakes
- gave the remainder instead of what still has to be added: 8
- added one divisor too many: 18
- added a whole divisor when less than that was needed: 13
Question 3
Add up the first 45 natural numbers.
-
A
2070
-
B
1035
-
C
2115
-
D
2025
Answer: Option B — with explanation
There is a standard result for each of the three series.
Sum = n(n + 1)/2 = 45 x 46 / 2 = 1035.
Common mistakes
- used the formula for the sum of the first n even numbers: 2070
- took the square of n, which is the sum of the first n odd numbers: 2025
- used the wrong product in the even-number formula: 2115
Question 4
What is the unit digit of $2^{14}$?
Answer: Option B — with explanation
The unit digit of powers of 2 repeats in a cycle, so only the position of 14 within that cycle matters.
The unit digits run through 2, 4, 8, 6, a cycle of length 4.
The power 14 sits at position 2 in that cycle, which is the digit 4.
Common mistakes
- counted one place too few round the cycle of unit digits: 3
- counted two places too far round the cycle: 6
- counted one place too far round the cycle of unit digits: 5
Question 5
Find the least number which, when added to 1837, gives a number divisible by 5.
Answer: Option A — with explanation
Dividing 1837 by 5 leaves a remainder of 2.
So 1837 is short of the next multiple by 5 - 2 = 3.
Check: 1837 + 3 = 1840, and 1840 / 5 = 368 exactly.
Common mistakes
- added a whole divisor when less than that was needed: 5
- added one divisor too many: 8
- gave the remainder instead of what still has to be added: 2
Frequently asked questions
How long is the cycle of unit digits?
It is at most four. It is four for the bases 2, 3, 7 and 8, two for 4 and 9, and one for 5 and 6.
What if the remainder is zero?
Then the power sits on the last entry of the cycle. For the base 7 the cycle is 7, 9, 3, 1, so a remainder of zero means the unit digit is 1.
How do I find the smallest number to add for divisibility?
Divide and find the remainder, then the answer is the divisor minus that remainder. If the remainder is 3 and the divisor is 7, add 4.
What is the sum of the first n odd numbers?
It is n squared. The first 10 odd numbers add up to 100, because 10 squared is 100.
