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

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There are 18 numbers with 2 digits that are divisible by 5. First 2 digit number is 10 → 10 ÷ 5 = 2 → first 2 digit number divisible by 25 is 5 × 2 Last 2 digit number is 99 → 99 ÷ 5 = 19 4/5 → last 2 digit number divisible by 5 is 5 × 19 → There are 19 - 2 + 1 = 18 numbers with 2 digit divisible by 5.

No one-digit numbers are divisible by 10. 10 is divisible by 1, 2 and 5.

Eight of them.

6 (15,30,45,60,75,90)

No. Numbers divisible by 2 have a digit in the ones place that is divisible by 2. In others words, numbers ending in 0,2,4,6, or 8

13

If n is divisible by both 5 and 6, then it should be divisible by 30 (5 * 6). Considering you are asking for only two-digit numbers, the answer(s) would be 30, 60, and 90. So, three numbers.

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.

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.

None. Extremely simple proof: To be divisible by 5 the units digit must be 5. But then it would not be divisible by 2.

Even numbers are divisible by two, and half of all numbers are even, so there is a 50 percent chance that a four-digit number is divisible by two.

All whole numbers are divisible by 1. Numbers are divisible by 2 if they end in 2, 4, 6, 8 or 0. Numbers are divisible by 3 if the sum of their digits is divisible by 3. Numbers are divisible by 4 if the last two digits of the number are divisible by 4. Numbers are divisible by 5 if the last digit of the number is either 5 or 0. Numbers are divisible by 6 if they are divisible by 2 and 3. Numbers are divisible by 9 if the sum of their digits is equal to 9 or a multiple of 9. Numbers are divisible by 10 if the last digit of the number is 0.

There are five numbers: 18, 36, 54, 72, 90.

There are 4,500 such numbers. One of them is 5646

Numbers that are co-prime won't both be divisible by nine.

A number which which ends with a 2 digit number that is divisible by 4. For example, 1028 ends with 28 which is divisible by 4.

All real numbers are divisible by 2. Only the even numbers (there are infinitely many) are divisible by 2 without remainder.

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.

48.The first two digit number is 10, the last two digit number is 99, so there are 99 - 10 + 1 = 90 two digit numbers→ 10 ÷ 3 = 31/3 → first two digit number divisible by 3 is 4 x 3 = 12→ 99 ÷ 3 = 33 → last two digit number divisible by 3 is 33 x 3 = 99→ 33 - 4 + 1 = 30 two digit numbers divisible by 3→ 10 ÷ 5 = 2 → first two digit number divisible by 5 is 2 x 5 = 10→ 99 ÷ 5 = 194/5 → last two digit number divisible by 5 is 19 x 5 = 95→ 19 - 2 + 1 = 18 two digit numbers divisible by 5→ 30 + 18 = 48 two digit numbers divisible by 3 or 5 OR BOTH.The numbers divisible by both are multiples of their lowest common multiple: lcm(3, 5) = 15, and have been counted twice, so need to be subtracted from the total→ 10 ÷ 15 = 010/15 → first two digit number divisible by 15 is 1 x 15 = 15→ 99 ÷ 15 = 69/15 → last two digit number divisible by 15 is 6 x 15 = 90→ 6 - 1 + 1 = 6 two digit numbers divisible by 15 (the lcm of 3 and 5)→ 48 - 6 = 42 two digit numbers divisible by 3 or 5.→ 90 - 42 = 48 two digit numbers divisible by neither 3 nor 5.

78,84,90,96

120 and 124, among others.

Three of them.

There are 6.They are:15,45,75,51,54,57.If you want to include negative numbers (i.e. -15 etc...) then you can double the answer.

between 1 and 600 inclusive there are:300 numbers divisible by 2200 numbers divisible by 3100 numbers divisible by both 2 and 3400 numbers divisible by 2 or 3.

Last digit (0) is even, so it is divisible by 2 4 + 3 + 2 + 0 = 9 which is divisible by 3, so it is divisible by 3 last digit is 0 or 5, so it is divisible by 5 4 + 3 + 2 + 0 = 9 which is divisible by 9, so it is divisible by 9 last digit is 0, so it is divisible by 10 → 4320 is divisible by all the numbers 2, 3, 5, 9, 10