Best Answer

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.

🙏

🤨

😮

Study guides

Q: How many two digit numbers divisible by neither 3 nor 5?

Write your answer...

Submit

Related questions

None. 3 digit numbers are not divisible by 19 digit numbers.

There are 151 3-digit numbers that are divisible by 6.

There are 600 5-digit numbers divisible by 150.

There are 180 3-digit numbers divisible by five.

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

18 numbers are there

There are 30 such numbers.

A three digit number cannot be divisible by a 5 digit number - in any base.

There are 282 such numbers.

30 of them.

13

2999 of them.

There are 300.

250 of them.

81,818,180 of them.

There are 278 5-digit numbers less than 20,000 that are divisible by both four and nine.

There are 90,000 such numbers.

There are 5! = 120 such numbers.

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.

-10

-16

-22

Eight of them.

There are 112 such numbers.

The first 3 digit number divisible by 19 is 114 (= 19 x 6) The last 3 digit number divisible by 19 is 988 (= 19 x 52) That means that there are 52 - 6 + 1 = 47 three digit numbers that are divisible by 19.