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If m is a natural number less than or equal to 100

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If m is a natural number less than or equal to 100  [#permalink]

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New post 25 Dec 2019, 02:54
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If m is a natural number less than or equal to 100, what is the probability that \((7^m + 11^m)\) is divisible by 5?

A. \(\frac{1}{4}\)

B. \(\frac{3}{7}\)

C. \(\frac{8}{25}\)

D. \(\frac{13}{50}\)

E. \(\frac{3}{8}\)
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Re: If m is a natural number less than or equal to 100  [#permalink]

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New post 25 Dec 2019, 07:14
last digit of 7^{4k+1} is 7
last digit of 7^{4k+2} is 9
last digit of 7^{4k+3} is 3
last digit of 7^{4k} is 1

Last digit of 11^m is always 1

m=4k+1, last digit of (7^m + 11^m)= 7+1=8( not divisible by 5}
m=4k+2, last digit of (7^m + 11^m)= 9+1=0( divisible by 5}
m=4k+3, last digit of (7^m + 11^m)= 3+1=4( not divisible by 5}
m=4k, last digit of (7^m + 11^m)= 1+1=2( not divisible by 5}

probability that \((7^m + 11^m)\) is divisible by 5, 0<m≤100= 1/4


uchihaitachi wrote:
If m is a natural number less than or equal to 100, what is the probability that \((7^m + 11^m)\) is divisible by 5?

A. \(\frac{1}{4}\)

B. \(\frac{3}{7}\)

C. \(\frac{8}{25}\)

D. \(\frac{13}{50}\)

E. \(\frac{3}{8}\)
Intern
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If m is a natural number less than or equal to 100  [#permalink]

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New post 29 Dec 2019, 03:59
We need to find a number which has last digit either 0 or 5 in order to be divisible with 5.

The last digit of 11^m will always be 1

The last digit of 7^m will be (7, 49, 343, 2401, 16807), this means that 7 has a cyclexity of 4. We are only interested in values where last digit is 9, so that the sum has last digit as 0 (9 + 1).

Therefore the total possible values of m (that are less than or equal to 100) will be 100/4 = 25.

The probability will be 25/100 or 1/4.
GMAT Club Bot
If m is a natural number less than or equal to 100   [#permalink] 29 Dec 2019, 03:59
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If m is a natural number less than or equal to 100

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