# How many 3 digit numbers are there which are divisible by 2,6and 10 but not divisible by 13,29?

How many 3 digit numbers are there which are divisible by 2,6 and 10 but not divisible by 13,29 ? I am trying this question by calculating the number of numbers which are divisible by 2, divisible by 6 and divisible by 10, divisible by 13, divisible by 29. So, here I have to use the formula $$T_n$$ $$=$$ $$a+(n-1)d$$, where $$T_n$$ is the last term of the A.P. , $$a$$= 1st term of the A.P. and $$d$$ is the common difference of the A.P. The no. of 3 digit numbers are 900. {100,101,...999} are the 3 digit numbers. Now the 3 digit numbers which are divisible by 2 are$$-$$ {100,102,...998}, divisible by 6 are $$-$$ {102,108,114,...,996}, divisible by 10 are $$-$$ {11,110,...990}, divisible by 13 are $$-$$ {104,117,...,988}, divisible by 29 are $$-$$ {116,145,...986}. But I can't find the number of numbers that are divisible by 2,6,10 but not divisible by 13,29. Please help me out.

• My previous question was wrong. So, I deleted the question. Actually one question of this type was asked to us in our university by professors. Commented Nov 3, 2023 at 16:26

The numbers divisible by 2, 6 and 10 are the same as the numbers divisible by LCM[2,6,10] = 30. These would be 120, 150, ...,990 totaling to $$\left\lfloor\frac{900}{30} \right \rfloor = 30$$ 3-digit numbers divisible by 2,6, and 10.
13 and 29 share no common divisors with 30. So for a number $$n$$ to also be divisible by 13 or 29 it would need to have either the form:
$$n = 13 * 30 * z = 390 z$$ or $$n = 29 * 30 * z = 870 z$$
for some positive integer $$z$$.