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Find the least number which when divided by 6,15 and 18 leave remainder 5 in each case.
Given:
6, 15 and 18.
To do:
We have to find the smallest number that when divided by 6, 15 and 18 leaves a remainder of 5 in each case.
Solution:
The smallest number that when divided by 6, 15 and 18 leaves a remainder of 0 in each case is the LCM of 6, 15 and 18.
So,
The smallest number that when divided by 6, 15 and 18 leaves a remainder of 5 in each case is $=$ LCM of 6, 15 and 18 $+$ 5
Now,
Finding LCM of 6, 15 and 18:
Writing all the numbers as a product of their prime factors:
Prime factorization of 6:
$6=2\times3$
Prime factorization of 15:
$15=3\times5$
Prime factorization of 18:
$18=2\times3\times3$
LCM of 6, 15 and 18 $=2\times3\times3\times5$
$=90$
Thus,
LCM$(6, 15, 18) = 90$
The smallest number that when divided by 6, 15 and 18 leaves a remainder of 5 in each case is $=$ LCM of 6, 15 and 18 $+$ 5
The least number that when divided by 6, 15 and 18 leaves a remainder of 5 in each case is $=90+5$
$=95$
The least number that when divided by 6, 15 and 18 leaves a remainder of 5 in each case is 95.
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