Method to solve

imo 6 is the answer

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

pradeepkaushal9518 wrote:imo 6 is the answer

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

Nope!

OA 222

did i make some mistake in understanding the question? anyone

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

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....

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

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!

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

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.