Tanmay thought of a two-digit number and divided the number by the sum

Question

Tanmay thought of a two-digit number and
divided the number by the sum of the digits
of the number. He found that the remainder
is 3. Sonal also thought of a two-digit
number and divided the number by the sum
of the digits of the number. She also found
that the remainder is 3. Find the probability
that the two-digit number thought by Tanmay
and Sonal is same.
a. 1/11
b. 1/12
c. 1/13
d. 1/14
e. 1/15

nick1816 · Answer

10x+y -3 = 0 mod (x+y)
9x+x+y-3 = 0 mod(x+y)
9x-3= 0 mod(x+y)

1. when x= 1; y=5 (As 18>x+y>3 and must be factor of 9*1-3=6)
2. when x=2; y=3 (As 18>x+y>3 and must be factor of 9*2-3=15)
3. when x=3; y=1,3,5 or 9 (As 18>x+y>3 and must be factor of 9*3-3=24)
4. when x=4; y=7 (As 18>x+y>=4 and must be factor of 9*4-3=33)
5. when x=5; y=1,2 or 9 (As 18>x+y>=5 and must be factor of 9*5-3=42)
6. when x=6; y=no values possible (As 18>x+y>=6 and must be factor of 9*6-3=51)
7. when x=7; y=3, 5 or 8 (As 18>x+y>=7 and must be factor of 9*7-3=60)
8. when x=8; y= no values possible (As 18>x+y>=8 and must be factor of 9*8-3=69)
9. when x= 9; y=13 (As 18>x+y>=9 and must be factor of 9*9-3=78)

Total 14 values are possible.

the probability that the two-digit number thought by Tanmay and Sonal is same = (1/14)

GMATGuruNY · Answer

Let the two-digit integer = 10T+U, where T = the tens digit and U = the units digit.
Dividing 10T+U by its digit sum must yield a remainder of 3.
Thus, 10T+U is equal to a MULTIPLE OF ITS DIGIT SUM + 3:
 10T+U = m(T+U)+3

\frac{10T+U-3}{T+U} = m

\frac{9T+T+U-3}{T+U} = m

\frac{9T-3}{T+U} + \frac{T+U}{T+U} = m

\frac{9T-3}{T+U} + 1 = m

Constraints:
Since m must be an integer, the divisor T+U must be a factor of 9T-3.
Since a divisor must be greater than a remainder, and the remainder here is 3, T+U > 3.
Thus, T+U must be A FACTOR OF 9T-3 THAT IS GREATER THAN 3.
The following cases and options are yielded:

T=1, with the result that 9T-3 = 6
Options for T+U that divide evenly into 6 -->  1+5=6

T=2, with the result that 9T-3 = 15
Options for T+U that divide evenly into 15 -->  2+3=5

T=3, with the result that 9T-3 = 24
Options for T+U that divide evenly into 24 --> 3+1=4, 3+3=6, 3+5=8, 3+9=12

T=4, with the result that 9T-3 = 33
Options for T+U that divide evenly into 33 -->  4+7=11

T=5, with the result that 9T-3 = 42
Options for T+U that divide evenly into 42 -->  5+1=6, 5+2=7, 5+9=14

T=6, with the result that 9T-3 = 51
Options for T+U that divide evenly into 51 --> NONE

T=7, with the result that 9T-3 = 60
Options for T+U that divide evenly into 60 -->  7+3=10, 7+5=12, 7+8=15

T=8, with the result that 9T-3 = 69
Options for T+U that divide evenly into 69 --> NONE

T=9, with the result that 9T-3 = 78
Options for T+U that divide evenly into 78 -->  9+4=13

The blue results above indicate the following:
Total number of options for T+U = 14

Thus:
Tammay can select any of the 14 options.
P(Sonal chooses the same option as Tammay)  = \frac{1}{14}

D

Rogermoris · Answer

Any two digit can be represented by 10x(ten's digit) + y(units digit)
Now, we have that the remainder when this two digit number is divided by sum (x+y) is 3. 
Thus, 10x+y -3 = 0 mod (x+y)
9x+x+y-3 = 0 mod(x+y)
9x-3= 0 mod(x+y)

For values of x=1,2,3...10 we find the corresponding value of 9 to find the range of values satisfying the condition.

With subsitutions we get 14 values. Thus probability that the two-digit number thought by Tanmay and Sonal is same = (1/14)

Tanmay thought of a two-digit number and divided the number by the sum

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