Math, asked by achaldps, 10 months ago

If 1461 is divisible by 6^n, then find the maximum value of n

Answers

Answered by ayushyadav143
0

Answer:

Your answer is given in the following steps-:::

If 146 is divisible by 6^n , then what is the maximum value of n?

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Let us consider the following equation:

146/(6^n) = x

If 146 is divisible by (6^n), the x has to be an integer, and not a decimal or a fraction.

The factor of 146 should be the same as of 6^n’s factors.

Factors of 146: 2,73

Factors of 6^n: 2^n, 3^n

As 73 is a prime number, only 1 and 73 are divisible to 73.

No multiple of 2 and 3 divide 73 perfectly.

Thus we definitely can get a value of n but it won’t be a positive integer.

Now because we want the maximum value of n, the term 6^n has to be maximum.

The maximum number 146 is divisible by is 146 itself.

Hence 146=(6^n)

By Logarithms,

log(146)=n log(6)

n=2.16435285578/0.77815125038

n=2.7814

Maximum value of n= 2.7814

Answered by Anonymous
0

Step-by-step explanation:

If 146 is divisible by 6^n , then what is the maximum value of n?

Let us consider the following equation:

146/(6^n) = x

If 146 is divisible by (6^n), the x has to be an integer, and not a decimal or a fraction.

The factor of 146 should be the same as of 6^n’s factors.

Factors of 146: 2,73

Factors of 6^n: 2^n, 3^n

As 73 is a prime number, only 1 and 73 are divisible to 73.

No multiple of 2 and 3 divide 73 perfectly.

Thus we definitely can get a value of n but it won’t be a positive integer.

Now because we want the maximum value of n, the term 6^n has to be maximum.

The maximum number 146 is divisible by is 146 itself.

Hence 146=(6^n)

By Logarithms,

log(146)=n log(6)

n=2.16435285578/0.77815125038

n=2.7814

Maximum value of n= 2.7814

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