Question

a number m is such that it is divided by 30,36 and 45 the remainder is always 7.find the smallest possible value of m

Answer 1

${\huge{\pink{\underline{\underline{\purple{\mathbb{\bold{\mathcal{AnSwEr..}}}}}}}}}$

▪︎ For this we have to first find the LCM of 30, 36 and 45.

$\implies36 = 2 \times 13 \\ \\ \implies30 = 2 \times 15 \\ \\ \implies45 = 5 \times 9$

${\large{\blue{\bold{\underline{Therefore,}}}}}$

$lcm = 2 \times 13 \times 15 \times 5 \times 9 \\ \\ = 17550$

${\large{\blue{\bold{\underline{So,}}}}}$

☆ Required number,

$= 17550 + 7 \\ \\ = 17557$

¤ Therefore the number is 17557