Question

Let a, b be positive integers such that 2000 divides aºb. The least value of ab is​

Answer 1

Answer:

2000= 2^4 • 5 ^3

Step-by-step explanation:

a = 4

b= 5

ab = 20

Answer 2

Complete question: Let a, and b be positive integers such that 2000 divides $a^{a} b^{b}$. What is the least value of ab?

Answer:

The least value of ab is 20.

Given,

a, and b be two positive integers such that 2000 divides $a^{a} b^{b}$.

To Find,

The least value of ab.

Solution,

The method of finding the least value of ab in the given problem is as follows -

Prime factorisation of 2000 is $2000=2^{4} *5^{3}$.

Also, $2000=2^{4} *5^{3} =4^{2} *5^{3}$.

Since 2000 divides $a^{a} b^{b}$, the factors of 2000 must be present in the prime factorisation of $a^{a} b^{b}$.

If we choose a=2 and b=5, $a^{a} b^{b}= 2^{2}*5^{5}$ which is not divisible by 2000.

But, if we choose a=4 and b=5, $a^{a} b^{b}= 4^{4}*5^{5}$ is divisible by 2000 and these are the least values of a and b which will satisfy the condition.

So, the least value of ab will be $4*5=20$.

Hence, the least value of ab is 20.

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