# What is the smallest positive integer which is divisible by 11 and sum of all digits equals 13?

*print*Print*list*Cite

Here is another approach to the problem:

The integer can not be a two digit one as all the digits add up to a non even sum.

Let the three digit integer be xyz.

Then, x+y+z=13 (given) --- (i)

Again, for multiples of 11, the difference of the sum of alternate numbers is either zero or, 11.

So, x+z-y=0, 11 --- (ii)

Solving equations (i) and (ii), when x+z-y=0,

`2(x+z)=13`

`rArr x+z=13/2` , but sum of two digits can not be a fraction.

Hence `x+z-y=11`

`rArr 2(x+z)=24`

`rArr x+z=12 `

Scanning for the lowest possible non zero values of x, when x=1

z=11 (not a single digit), discarded.

when x=2, z=10 (not a single digit), discarded.

When x=3, z=9

Put these values of x and z in eq.(i) to get y=1

**Therefore, the smallest possible integer which is multiple of 11 and have the sum of all digits equalto 13 is 319.**

The smallest positive integer divisible by 11 and having sum of integers equal to 13 has to be found out.

Scanning through the multiples of 11 (smallest first), the two digit integers are all symmetrical figures, hence have the sum of all digits as an even number.

For three digit multiples of 11, sum of all the digits is given within parentheses:

11*10=110 (2), 11*11=121 (4), 11*12=132 (6), 11*13=143 (8), 11*14=154 (10), 11*15=165 (12), 11*16=176 (14), 11*17=187 (16), 11*18=198 (18), 11*19=209 (11), 11*20=220 (4), 11*21=231 (6), 11*22=242 (8), 11*23=253 (10), 11*24=264 (12), 11*25=275 (14), 11*26=286 (16), 11*27=297 (18), 11*28=308 (11), 11*29=319 (13).

**Therefore, the smallest positive integer divisible by 11 and having sum of its digits equal to 13 is 319.**