Find the greatest number of five digits which | vedclass

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

Question

(D) First, observe the difference between the divisors and their respective remainders:$12 - 5 = 7$$15 - 8 = 7$$21 - 14 = 7$$25 - 18 = 7$$28 - 21 = 7$

Vedclass · Accepted Answer

$98693$

Vedclass · Answer

(D) First, observe the difference between the divisors and their respective remainders:$12 - 5 = 7$$15 - 8 = 7$$21 - 14 = 7$$25 - 18 = 7$$28 - 21 = 7$Since the difference is constant $(7)$, the required number is $(L.C.M. 	ext{ of } 12, 15, 21, 25, 28) 	imes k - 7$.Calculate the $L.C.M.$ of $12, 15, 21, 25, 28$:$12 = 2^2 	imes 3$$15 = 3 	imes 5$$21 = 3 	imes 7$$25 = 5^2$$28 = 2^2 	imes 7$$L.C.M. = 2^2 	imes 3 	imes 5^2 	imes 7 = 4 	imes 3 	imes 25 	imes 7 = 2100$.The greatest five-digit number is $99999$.Divide $99999$ by $2100$:$99999 \div 2100 = 47$ with a remainder of $1299$.Subtract the remainder from $99999$ to find the largest multiple of $2100$ less than $100000$:$99999 - 1299 = 98700$.Finally, subtract the constant difference $(7)$: $98700 - 7 = 98693$.

Find the greatest number of five digits which when divided by $12, 15, 21, 25$ and $28$ leaves $5, 8, 14, 18$ and $21$ as remainders, respectively.

Similar Questions

Two numbers,both greater than $29$,have $HCF$ $29$ and $LCM$ $4147$. The sum of the numbers is

Find the least number of five digits which when divided by $12, 16, 21, 36$ and $40$ leaves a remainder of $8$ in each case.

What is the least value that should be added to $1812$ to make it divisible by $7, 11,$ and $14$?

Find the greatest number which will divide $25, 73$ and $97$ so as to leave the same remainder in each case.

The largest natural number which exactly divides the product of any four consecutive natural numbers is

Vedclass Test Series

Exam Paper Generator

Online Exam Module