# There are two-digit numbers such that the sum of these digits is 6 while the product of the digits is 1/3 of the original number. Find the numbers.

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Find all numbers of the form 10p+q, p and q nonnegative integers, such that p+q=6 and `pq=1/3(10p+q)` :

We can write all numbers of the form 10p+q such that p+q=6:

15,24,33,42,51

Now `1*5=1/3(15)`

`2*4=1/3(24)`

`3*3!=1/3(33)`

`4*2!=1/3(42)`

`5*1!=1/3(51)`

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**The numbers we seek are 15 and 24**

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here let the digit of ten's place be x and that of one's place be y. The the no. can be written in the form 10x+y. for eg: 91 can be given as 10*9+1=90+1=91.

now, x+y=6 and xy=1/3(10x+y)

or, xy=10x/3+y/3

or,xy-10x/3=y/3

or, x(y-10/3)=y/3

or, x=(y/3)/(y-10/3)

put value of x in x+y=6

then solve and you get,

(y-4)(y-5)=0

Thus either y=5 OR y=4

when y=5, x=6-5=1

when y=6, x=6-4=2

Thus one no.=10*1+5=15

another no. =10*2+4=24

Therefore, the two numbers are 15 and 24.