If the integer s m and n are chosen at random between $$1$$ and $$100$$, then the probability that a number of the form $$7m+7n$$ is divisible by $$5$$ equals

Question

If the integer s m and n are chosen at random between $$1$$ and $$100$$, then the probability that a number of the form $$7m+7n$$ is divisible by $$5$$ equals

Infinity Learn · Accepted Answer

We know that $$71=7$$,$$72=49$$,$$73=343$$,$$74=2401$$,$$75=16807$$. Therefore $$7k(where$$ k∈$$Z)$$ results in a number whose unit's digit is $$7$$ or $$9$$ or $$3$$ or $$1$$. Now $$7m+7n$$  will be divisible by $$5$$ if unit's place digit in the resulting number is $$5$$ or $$0$$. Clearly, it can never be $$5$$. But it can be $$0$$ if we consider values of m and n such that the sum of unit's place digits become $$0$$. And this can be done by $$choosingm=1$$,$$5$$,$$9$$,.... $$97n=3$$,$$7$$,$$11$$,...., $$99$$ (25$$ options $$each) [$$7+3=10$$]$$orm=2$$, $$6$$, $$10$$, ....,   $$98n=4$$,  $$8$$, $$12$$,...., $$100$$ (25$$ options $$each) [$$9+3=10$$]Therefore, the total number of selections of m, n such that $$7m+7n$$ is  divisible by $$5$$ is $$25$$×$$25+25$$×$$25100$$×$$100=14$$

If the integer s m and n are chosen at random between 1 and 100, then the probability that a number of the form $7^{m} + 7^{n}$ is divisible by 5 equals

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If the integer s m and n are chosen at random between 1 and 100, then the probability that a number of the form 7m+7n is divisible by 5 equals

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If the integer s m and n are chosen at random between 1 and 100, then the probability that a number of the form $7^{m} + 7^{n}$ is divisible by 5 equals