Given that n = 10^a + 10^b + 10^c, where a, b, and c are distinct

Question

Given that  n = 10^a + 10^b + 10^c , where a, b, and c are distinct positive integers, how many different positive values of n result if n is less than 1 billion (1,000,000,000) ?

A. 28
B. 36
C. 56
D. 84
E. 120

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chetan2u · Accepted Answer

What will be the minimum value of any of the variable a, b and c? :  It will be 1, as all are positive integers. 
What will be the maximum value of any of the variable a, b and c? :   It will be 8, as  10^9  will make n>1 billion. 
If the maximum of three is 8, then other 2 can be taken in 7C2 or 21 ways.
If the maximum of three is 7, then other 2 can be taken in 6C2 or 15 ways.
If the maximum of three is 6, then other 2 can be taken in 5C2 or 10 ways.
If the maximum of three is 5, then other 2 can be taken in 4C2 or 6 ways.
If the maximum of three is 4, then other 2 can be taken in 3C2 or 3 ways.
If the maximum of three is 3, then other 2 can be taken in 2C2 or 1 way.

Each of the way will give us a different value of n, so 21+15+10++3+1 or 56 ways.

Blair15 · Answer

There are 8*7*6 ( We start from 1 as only positive integers have to be considered, and we end at 8 since 10^9 would give 1 billion) ways of getting different combinations. But since the order of the numbers does not matter, we have to remove the repetitive cases by dividing 8*7*6 by 3! ( there are 3! ways of reshuffling the numbers).
Therefore - 8*7*6/3! = 56.

C  is the  right answer  IMO.

Karnk · Answer

In this maximum power of 10 can be 8. coz 10^9 would result in the number n >10^9. so powers of 10 can be anywhere from 1 to 8. and we need to pick 3 out of 8 in dfrnt ways. so various values of n = 8c3 = 56.

adewale223 · Answer

Maximum power of 10 in the arrangement is 8, minimum power is 1

Therefore

8 x 7 x 6/3!

8x 7

= 56

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Akshaynandurkar · Answer

Method 1:

This is a selection problem where we are selecting three distinct numbers from 1 to 8 (8 nos) which can be selected in 8C3 ways = 56

Method 2:

From above question, it can be conclude that maximum of a,b,c cannot be more than 8
consider, one of a,b,c as 1, then we have following sets considering one of the three as 1
123,124,125,126,127,128(Total 6)
134,135,136,137,138 (Total 5)
145,146,147,148 (Total 4)
156,157,158(Total 3)
167,168 (Total 2)
178(total 1)

So if we consider one of the value as 1 then we get total combinations of a,b and c = 6+5+4+3+2+1 = 21

Similarly consider one of the values as 2, then we get following combinations
234,235,236,237,238 (Total 5) (Note 213 cannot be considered here because it was already covered with 123)
245,246,247,248 (Total 4)
Total 3
Total 2
Total 1
Summation = 5+4+3+2+1 = 15

Here you can see the pattern 
next summation will be = 4+3+2+1 = 10 (with 3)
next summation will be = 3+2+1 = 6 (with 4)
next summation will be = 2+1 = 3 (with 5)
next summation will be = 1 (with 6)

Grand summation = 21+15+10+6+3+1= 56

Given that n = 10^a + 10^b + 10^c, where a, b, and c are distinct

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GMAT Problem Solving (PS) Questions

Given that n = 10^a + 10^b + 10^c, where a, b, and c are distinct

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