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What I've done here is - take the unit's digit of all the individual numbers into consideration because the unit's digit of the sum of the digits will be decided ultimately by the unit's digit of all the individual numbers.

Thereafter, if you see the expression, I've grouped together positive and negative numbers.

6^15 is a very big number and much > 7^4 + 9^3.

So you need to visualize it as (xxxx6 - 10), and 6 can borrow 1 easily from left side because it's very big number.

HTH. _________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

As subhashgosh did, let me just add visual to it. Interesting to see that 6^15 is large number something like XXXXXXXX6 7^4 = 2401 9^3 = 729 Now subtract XXXXXXXX6 - 2401 -------------------------- xxxxxxxxxx5

Re: What is the units digit of 6^15 - 7^4 - 9^3? [#permalink]
21 May 2012, 19:12

1

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Ok Let me try to explain this.

If you need to subtract 7 from 234 then as you know 7 is greater than 4 so you will take one carry from 10th place integer and you will subtract 7 from 14 to get 7. So result of this subtraction is 227.

Same way we know the last digit of the expression is 6 but this less than 10 so we have take carry from 10th place. So it becomes 16.

All numbers from 1-9 repeat their units digit after a cycle of 4. i.e. 3^2 will have same units digit as 3^6 and form above table we see that 9 would be that digit.

Similarly, 6^15 will have units digit=6 7^4 will have units digit=1 9^3 will have units digit=9

Re: What is the units digit of 6^15 - 7^4 - 9^3? [#permalink]
22 May 2012, 03:36

Expert's post

also you can see in this way:

6-9 = -4 but our six is a number in our equation that end with unit digit 5 infact to obtain the number of our answer choices subtracting 9 we have a number that must with unit digit of 5

A) 8 -> to get 8 you have 17 - 9= 8 but unit digit is 7 wrong

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