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All 2 digit numbers are divisible by 7: it is simply that some of the quotients are not whole numbers.

10/7 = 1 3/7

99/7 = 14 1/7

So the number of 2-digit numbers which are evenly divisible by 7 are 14-1 = 13.

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8y ago
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8y ago

The first one is 14, the last one is 98. Since every 7th. number is divisible by seven, you can take the difference, divide that by 7, and add 1 to the result (because of the endpoints).

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Q: How many 2 digit numbers are divisible by 7?
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Starting at 12 and ending at 99, there are 30 two-digit numbers divisible by three.


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100. The highest 3 digit number divisible by 9 is 999 (111 times). The highest 2 digit number divisible by 9 is 99 (11 times). The difference is 100.


How many positive three digit integers are divisible by neither 2 nor 3?

There are 300 three digit numbers that are divisible by neither 2 nor 3. There are 999 - 100 + 1 = 900 three digit numbers. 100 ÷ 2 = 50 → first three digit number divisible by 2 is 50 × 2 =100 999 ÷ 2 = 499 r 1 → last three digit number divisible by 2 is 499 × 2 = 998 → there are 499 - 50 + 1 = 450 three digit numbers divisible by 2. 100 ÷ 3 = 33 r 1 → first three digit number divisible by 3 is 34 × 3 = 102 999 ÷ 3 = 333 → last three digit number divisible by 3 is 333 × 3 = 999 → there are 333 - 34 + 1 = 300 three digit numbers divisible by 3. The lowest common multiple of 2 and 3 is 6, so these have been counted in both those divisible by 2 and those divisible by 3. 100 ÷ 6 = 16 r 4 → first three digit number divisible by 6 is 17 × 6 = 102 999 ÷ 6 = 166 r 3 → last three digit number divisible by 6 is 166 × 6 = 996 → there are 166 - 17 + 1 = 150 → there are 450 + 300 - 150 = 600 three digit numbers that are divisible by either 2 or 3 (or both). → there are 900 - 600 = 300 three digit numbers that are divisible by neither 2 nor 3.

Related questions

How many 2 digit numbers are divisible by 3?

Starting at 12 and ending at 99, there are 30 two-digit numbers divisible by three.


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