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.
$\begingroup$
$\endgroup$
1
asked Nov 3, 2023 at 16:06
-
$\begingroup$ 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. $\endgroup$– Syamaprasad ChakrabartiCommented Nov 3, 2023 at 16:26
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
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 $z$.
So the numbers satisifying this condition are 390, 780, and 870. Therefore, there are 30-3 = 27 3 digit positive integers which are divisible by 2,6 and 10 but not divisible by 13 or 29
