Question
For a positive integer n, let Pn denote the product of the digits of n and Sn denote the sum of the digits of n. The number of integers between 10 and 1000 for which Pn+Sn=n is (CAT 2005)
Answer: Option A
:
A
Sol:
(a)10<n<1000
Let n is two digit number
n=10a+b^a†′Pn=ab,Sn=a+b
then, ab+a+b=10a+b
^a†′ab=9a^a†′b=9
There are 9 such numbers 19,29,33,....,99.
Then let n is three digit number
^a†′n=100a+10b+c^a†′Pn=abc,Sn=a+b+c
Then abc+a+b+c=100a+10b+c
^a†′abc=99a+9b
^a†′bc=99+9
But the maximum value for bc=81
And RHS is more than 99. Hence, no such number is possible.
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:
A
Sol:
(a)10<n<1000
Let n is two digit number
n=10a+b^a†′Pn=ab,Sn=a+b
then, ab+a+b=10a+b
^a†′ab=9a^a†′b=9
There are 9 such numbers 19,29,33,....,99.
Then let n is three digit number
^a†′n=100a+10b+c^a†′Pn=abc,Sn=a+b+c
Then abc+a+b+c=100a+10b+c
^a†′abc=99a+9b
^a†′bc=99+9
But the maximum value for bc=81
And RHS is more than 99. Hence, no such number is possible.
Was this answer helpful ?
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