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by ssgmatter » Sat Jun 19, 2010 7:34 pm
If x represents the sum of all the positive three-digit numbers that can be
constructed using each of the distinct nonzero digits a, b, and c exactly once,
what is the largest integer by which x must be divisible?
3 6 11 22 222
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by pradeepkaushal9518 » Sat Jun 19, 2010 7:53 pm
imo 6 is the answer

7,8,9 is the 3 digit sum = 24
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by ssgmatter » Sat Jun 19, 2010 7:54 pm
pradeepkaushal9518 wrote:imo 6 is the answer

7,8,9 is the 3 digit sum = 24
Nope!

OA 222
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by pradeepkaushal9518 » Sat Jun 19, 2010 7:56 pm
did i make some mistake in understanding the question? anyone
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by selango » Sat Jun 19, 2010 9:29 pm
X represents sum of all the positive three-digit numbers that can be
constructed using each of the distinct nonzero digits in form of abc

it means abc=

123

132

134

143

sum of all 3 digits number that can be formed from 1,2.... 9 in abc
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by ssgmatter » Sat Jun 19, 2010 10:01 pm
selango wrote:X represents sum of all the positive three-digit numbers that can be
constructed using each of the distinct nonzero digits in form of abc

it means abc=

123

132

134

143

sum of all 3 digits number that can be formed from 1,2.... 9 in abc
Please explain in details....
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by amising6 » Sat Jun 19, 2010 10:50 pm
ssgmatter wrote:If x represents the sum of all the positive three-digit numbers that can be
constructed using each of the distinct nonzero digits a, b, and c exactly once,
what is the largest integer by which x must be divisible?
3 6 11 22 222
sum of all 3 digit number using 1,2,3,....9
111*(3-1)!(1+2+3+4+5+6+7+8+9)=111*4*45= so now out of the given option (now one concept to find the sum of n digit using 1 to n digit is given by 11111...n(n-1)!(1+2+3+..+n))111*4*45
111*2*2*3*3*5
will be divisible by 3,6,222 so largest wil be 222
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by ssgmatter » Sat Jun 19, 2010 11:24 pm
amising6 wrote:
ssgmatter wrote:If x represents the sum of all the positive three-digit numbers that can be
constructed using each of the distinct nonzero digits a, b, and c exactly once,
what is the largest integer by which x must be divisible?
3 6 11 22 222
sum of all 3 digit number using 1,2,3,....9
111*(3-1)!(1+2+3+4+5+6+7+8+9)=111*4*45= so now out of the given option (now one concept to find the sum of n digit using 1 to n digit is given by 11111...n(n-1)!(1+2+3+..+n))111*4*45
111*2*2*3*3*5
will be divisible by 3,6,222 so largest wil be 222
can you please explain how did you use this formula to solve this question???....Is this a GMAT type question??...

Thanks!
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by jube » Sun Jun 20, 2010 4:00 am
For a, b & c, the possible 3 digit nos. can be:

100a + 10b + c
100a + 10c + b
100b + 10a + c
100b + 10c + a
100c + 10b + a
100c + 10a + b

Summing up the 6 nos. that are possible: 100(2a + 2b + 2c) + 10(2a + 2b + 2c) + (2a + 2b + 2c)
=(2a + 2b + 2c )(100 + 10 +1)=2(a + b +c)111=222(a+b+c)

therefore the no. will always be divisible by 222. Hence, E
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by rohit56 » Mon Dec 17, 2018 10:14 pm
A 3-digit number 'abc' could be written as 100a + 10b + c.
Total possible arrangements of 'abc' are 3! = 6 arrangements
For exactly 2 times each digit a, b and c would be at each of the three places.
So, Sum of digits at hundredth place = 2(a + b + c)
Sum of digits at tens place = 2(a + b + c). Similar for ones place.
So, the total sum becomes = 100 x 2(a + b + c) + 10 x 2(a + b + c) + 2(a + b + c) = 222(a + b + c).
Hence, 222 is the answer.
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