Number 2

Let N N be a 3-digit positive integer and let M M be another 3-digit positive integer obtained by reversing the digits of N N .

Find the number of possible values of N N such that N M = 396 N-M= 396 .


The answer is 50.

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1 solution

Soumya Shrivastva
May 15, 2016

L e t N = 100 x + 10 y + z , S o M = 100 z + 10 y + x N M = 99 ( x z ) = 396 x z = 4 , S o p o s s i b l e v a l u e s o f ( x , z ) ( 5 , 1 ) , ( 6 , 2 ) , ( 7 , 3 ) , ( 8 , 4 ) & ( 9 , 5 ) N o w f o r a n y v a l u e o f ( x , y ) t h e r e a r e 10 v a l u e s o f y ( 0 9 ) H e n c e t h e r e a r e 5 × 10 = 50 p o s s i b l e v a l u e s . Let\quad N=100x+10y+z,\quad So\quad M=100z+10y+x\\ \Rightarrow N-M=99(x-z)=396\\ x-z=4,\quad So\quad possible\quad values\quad of\quad (x,z)\rightarrow (5,1),(6,2),(7,3),(8,4)\& (9,5)\\ Now\quad for\quad any\quad value\quad of\quad (x,y)\quad there\quad are\quad 10\quad values\quad of\quad y\quad (0-9)\\ Hence\quad there\quad are\quad 5\times 10=50\quad possible\quad values.

What about (4, 0)? Though you said that m and n are natural numbers, you didn't say anything about x, y, or z.

Sharon Matsuoka - 5 years, 1 month ago

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if we take x=4 & z=0,then N=4y0 but M=0y4. Which is not a 3-digit natural number...

Soumya Shrivastva - 5 years, 1 month ago

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