Find the least number which when divided by 1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) First,find the $L.C.M.$ of $16, 18, 20$ and $25$.$16 = 2^4$$18 = 2 	imes 3^2$$20 = 2^2 	imes 5$$25 = 5^2$$L.C.M. = 2^4 	imes 3^2 	imes 5^2 = 1

Vedclass · Accepted Answer

$18004$

Vedclass · Answer

(C) First,find the $L.C.M.$ of $16, 18, 20$ and $25$.$16 = 2^4$$18 = 2 	imes 3^2$$20 = 2^2 	imes 5$$25 = 5^2$$L.C.M. = 2^4 	imes 3^2 	imes 5^2 = 16 	imes 9 	imes 25 = 3600$.Any number that leaves a remainder of $4$ when divided by $16, 18, 20$ and $25$ is of the form $(3600K + 4)$,where $K$ is a natural number.We are given that this number is divisible by $7$.$(3600K + 4) \div 7 = 0$ remainder.$3600 \div 7$ leaves a remainder of $2$ (since $3600 = 7 	imes 514 + 2$).So,$(2K + 4)$ must be divisible by $7$.For $K = 1, 2K+4 = 6$ (not divisible by $7$).For $K = 2, 2K+4 = 8$ (not divisible by $7$).For $K = 3, 2K+4 = 10$ (not divisible by $7$).For $K = 4, 2K+4 = 12$ (not divisible by $7$).For $K = 5, 2K+4 = 14$ (divisible by $7$).Thus,the required number $= 3600 	imes 5 + 4 = 18000 + 4 = 18004$.

Find the least number which when divided by $16, 18, 20$ and $25$ leaves $4$ as a remainder in each case,but when divided by $7$ leaves no remainder.

Similar Questions

The $HCF$ of $x^{8}-1$ and $x^{4}+2x^{3}-2x-1$ is

The product of two numbers is $2028$ and their $HCF$ is $13$. The number of such pairs is

Find the greatest number which will divide $2112$ and $2792$ leaving the remainder $4$ in each case.

The traffic lights at three different road crossings change after every $48\, seconds,$ $72\, seconds$ and $108\, seconds,$ respectively. If they all change simultaneously at $8: 20: 00$ $hours,$ then they will again change simultaneously at

The least multiple of $7,$ which leaves a remainder of $4,$ when divided by $6, 9, 15,$ and $18$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module