Multiplying and dividing by two digits; divisibility · Usually met around P4
Divisibility rules for 4, 6, 8, 9, 25 and 100
4, 25, 100: last two digits. 8: last three. 9: the digit sum. 6: both 2 and 3.
How to do it
Method
4, 25, 100: last two digits. 8: last three. 9: the digit sum. 6: both 2 and 3
7335: 7 + 3 + 3 + 5 = 18, so divisible by 9
Another way
Divide to check
7335 ÷ 9 = 815, no remainder
In full
100 divides exactly by 4 and by 25, so the whole hundreds don't matter: look only at the last two digits. If they are divisible by 4, so is the number (7316: 16 ÷ 4 = 4). If they are 00, 25, 50 or 75, it is divisible by 25. If they are 00, it is divisible by 100. 1000 divides exactly by 8, so for 8 look at the last three digits (5120: 120 ÷ 8 = 15). For 9, add the digits: 7335 gives 7 + 3 + 3 + 5 = 18, a multiple of 9, so 7335 is divisible by 9. For 6, it must be divisible by both 2 and 3: even, with a digit sum that is a multiple of 3. Another way: divide once and see if anything is left: 7335 ÷ 9 = 815.
Worked examples
Have a go first, then open each one for the answer and the working.
Example 1
Which of these numbers is divisible by 25?
- A. 3995
- B. 8665
- C. 3900
- D. 4740
Choose the right answer
Show the answer and working
Answer: C. 3900
100 is 4 times 25, so look at the last two digits: only 00, 25, 50 and 75 divide by 25.
Example 2
The four-digit number 348□ is divisible by 4. What is the biggest digit □ can be?
Show the answer and working
Answer: 8
Only the last two digits matter: trying down from 9, 88 is the biggest that divides by 4, so □ is 8.
Example 3
How many of 4, 6, 8 and 9 divide 3036 exactly?
Show the answer and working
Answer: 2
Check each: 4, the last two digits; 8, the last three; 9, the digit sum; 6, even with a digit sum that is a multiple of 3. That makes 2.
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