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# Which of the following represents the largest 4 digit number which can be added 7249 in order to make the derived number divisible by each of 12,14,21,33 and 54.a)9123       b)9383c)8727       d)none of these

kartheeek
24 Points
6 years ago
ans is none of these
l.c.m. of 12,14,21,33,54 is 756
and 756*2=1512 is also divisible by all the numbers
so the ans is 1512+7249=8761

Lovesh Sinha
18 Points
6 years ago
LCM is wrong....check it again?

kartheeek
24 Points
6 years ago
ans is l.c.m.of 12,14,21,33,54 – 7249=8316-7429 or 8316*2-7429
Miriyla Gowtham
15 Points
3 years ago
THE LCM OF 12,14,21,33,54 IS 8316.
BUT 8316*2 = 16632 IS DIVISIBLE BY ALL ABOVE NUMBERS SO ANS IS 16632 – 7249 = 9383 (ANS)

nikhil
13 Points
2 years ago
Find the LCM of the given numbers.
12= 2*2*3
14= 2*7
21= 3*7
33= 3*11
54= 3*2*3*3

Max powers of all prime numbers:
2=2
3=3
7=1
11=1

LCM=2^2*3^3*7^1*11^1=8316

8316-7249=1067. Thus, if we add 1067 to 7249, the number will be divisible by all the given numbers.
However, 1067 is not the GREATEST 4-digit number to satisfy the condition.

Next number that will be divisible is:
1067+8316=9383

Thus, adding 9383 to 7249 will give us a number that will be divisible by all the given numbers PLUS 9383 is the greatest 4-digit number that satisfies this condition.