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Re: The units digit of 35^87 + 93^46 is: [#permalink]
17 Jul 2013, 09:55

1

This post received KUDOS

Step 1: (35)^(87) can be broken up into (5)^87 x (7)^87. Each power of 7 ends in a units digit of either a 7,9,3 or 1. Each power of 5 ends in a 5. When you multiply 5 by any odd number you will end up with a units digit of 5.

Step 2: (93)^(46) can be broken up into (3)^46 x (31)^46. Each power of 31 ends in a units digit of 1. Each power of 3 ends in a units digit of either 3,9,7 and 1. Since there is a pattern here where every 4th power of 3 ends in a units digit of 3, the 46th power of 3 would end in a units digit of 9. When you multiply 1 by 9 you end up with a units digit of 9.

Therefore, the units digit we are looking for is 5 + 9 = 14. Units digit will be 4. Answer B.

It is quite intuitive to go for a basic two step approach for this problem.

When dealing with 35^87, we can apply a simple concept here. 5 raised to any power > 0 must have 5 as its units digit. When dealing with 93^46, we can apply the concept of cyclicity. Since the cyclicity of 3 is 4, so units digit of 93^46 is equivalent to teh units digit of 3^2 i.e. 9. On adding these 2 digits i.e. 9 and 5, we get 14 of which the units digit is 4. Will be curious to know how others deal with such questions. _________________

It is quite intuitive to go for a basic two step approach for this problem.

When dealing with 35^87, we can apply a simple concept here. 5 raised to any power > 0 must have 5 as its units digit. When dealing with 93^46, we can apply the concept of cyclicity. Since the cyclicity of 3 is 4, so units digit of 93^46 is equivalent to teh units digit of 3^2 i.e. 9. On adding these 2 digits i.e. 9 and 5, we get 14 of which the units digit is 4. Will be curious to know how others deal with such questions.

Good question, I have the same question in my mind.

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