If a three-digit positive integer has its digits reversed, the

100x+10y+z-100z+10y+x=297
99(x-z)=297
x-z = 3
there are 6 possible numbers for x. for example: 9y6-6y9=297 and so on till x=4
as for y. there are 10 possibilities: from 0 to 9
so 6*10=60 pairs possible.
so answer is D

Ans: D

N=100x+10y+z
=> (100x+10y+z) - (100x+10y+z) =297
=> x-z =3
(1) x <=9, z <=6 <=> Total 6 pairs z=1,2,3,4,5 & 6
(2) y can be 0 to 9(Total 10)

Total possibilities = 6*10=60

What about a?
b*c= 10 * 6= 60
so what about a? where have it gone?

No need to consider a - above solution is self explanatory

Answer is D
Possible combinations are 4b1, 5b2, 6b3,7b4,8b5, 9b6
B can be any dogit from 0-9 , hence 10 possiblities. and the above combination are only 6 , therefore answer is 6*10 = 60

xyz-zyx=297
297/99=3=x-z
6 possible x-z combinations: 9-6,8-5,7-4,6-3,5-2,4-1
10 possible y values: 0-9
6*10=60 possible pairs
D

Par of GMAT CLUB'S New Year's Quantitative Challenge Set

Asked: If a three-digit positive integer has its digits reversed, the resulting three-digit positive integer is less than the original integer by 297. How many such pairs are possible?

Let the 3-digit positive integer be of the form xyz

(100x + 10y + z) - (100z + 10y + x) = 99(x-z) = 297
x -z = 3

Since x & z are not 0

(x,z) = {(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)} = 6 cases
y can take 10 values

Total such numbers = 6*10 = 60

IMO D

\( ABC-CBA=297:100a+10b+c-100c-10b-a=297, 99(a-c)=297, (a-c)=3\)
\((A,C)=(9,6;8,5;7,4;6,3;5,2;4,1)…(A,C)≠(3,0):C>0\)
\((A,C)=6.pairs…B=(0,1,2,3,4,5,6,7,8,9)=10\)
\(Total.cases:6*10=60\)

Ans (D)

Considering the three-digit number to be ABC.
The value of the three-digit number is given by 100*A+10*B+C.
When the digits are reversed the number becomes CBA.
The value becomes : 100*C + 10*B+A.
Since the value decreased by 297 this can be represented as :
\(\left(100A+10B+C\right)-\left(100C+10B+A\right)\ =\ 297\)
99(A-C) = 297.
A-C = 3.
The different possibilities include : (A = 9, C = 6), (A = 8, C = 5), (A = 7, C = 4), (A = 6, C = 3), (A = 5, C = 2), (A = 4, C = 1)
A and C cannot take the value of 0 because then the numbers do not satisfy the three digit number condition.
For the 6 case B can take any value from 0 to 9.
A total of 10*6 = 60 cases.

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