Find greatest 4 digit number which becomes divisible by 24 and 28 when 100 is added to it
Answers
Answer:
hey mate
here is ur answer..
Consider the LCM — the least common multiple. The LCM of 24 and 28 is the smallest number that is also divisible by 24 and 28. So anything divisible by both 24 and 28 must also be divisible by the LCM of 24, and 28.
To find the LCM of 24 and 28, we compute the prime factorization of each:
24 = 2^3 * 3
28 = 2^2 * 7
Next, we consider each prime, and take the largest exponent.
LCM(24, 28) = 2^3 * 3 * 7
Basically, anything divisible by both 24 and 28 must be divisible by both 2^3 *3 and 2^2 * 7, so it needs to have as factors 2^3, 3, and 7.
LCM(24, 28) = 168
Now we’re getting somewhere! The maximum value we’re going to look at is the biggest 4-digit number (9999) plus 100. This is 10099. So we’re looking for the biggest integer <= 10099 that is divisible by 168
Looking_for = floor(10099/168)
10099/168 = 60.113 (give or take), so floor(10099/168) is 60.
So the biggest thing <= 10099 that is divisible by 168 is 60*168.
60*168 is 10,080.
Subtract 100 from this and you have 9980.
There, a little math, a little use of a calculator, and we’re done!
hope it helps
Concept:
LCM ( least common multiple) of two numbers gives the smallest positive integer that is divisible by both numbers.
Given:
Two numbers 24 and 28.
Find:
We have to find the greatest 4-digit number which becomes divisible by 24 and 28 when 100 is added to it.
Solution:
For the LCM of 24 and 28, the prime factorization of each:
24 = 2³ ×3
28 = 2² ×7
LCM(24, 28) = 2³ × 3 × 7
= 168
By multiplying 168 with 59 we get the highest four-digit number as 9912.
We need a number which is 100 less than 9912.
Hence, 9812 is the greatest 4-digit number which becomes divisible by 24 and 28 when 100 is added to it.
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