Asked by Madhurima Mishra
when a 3 digits number is reversed, the number decreases by 396. the diffrence between the digits at the units place and the digit at the tens place is the same as the difference between the digits at the tens place and the hundreds place. if the product of 3 digits number is non zero then how many such 3 digits numbers are possible
Answers
Answered by
Steve
100u+10t+h = 100h+10t+u - 396
u-t = t-h
solve that and you get
t = h-2
u = h-4
h = t+2 = u+4
since no digit is zero, we can have
531
642
753
864
975
u-t = t-h
solve that and you get
t = h-2
u = h-4
h = t+2 = u+4
since no digit is zero, we can have
531
642
753
864
975
Answered by
Reiny
let the hundreds, tens , and unit digits be
x , y, and z respectively .
so the original number is
100x + 10y + z
the number reversed is
100z + 10y + x
100x + 10y + z - (100z + 10y + x) = 396
99x - 99z = 396
x - z = 4
x = z+4
z - y = y - x
2y = z+x
y = (z+x)/2
" if the product of 3 digits number is non zero "
What does that mean??
xyz ≠ 0 ???
cases:
z = 0, x = 4 , y = 2
number is 420 , but the product is zero
z = 1 , x = 5, y = 3
number is 531
z = 2 , x = 6, y = 4
number is 642
z = 3 , x = 7, y = 5
number is 753
z = 4, x = 8 , y = 6
number is 864
z = 5, x = 9, y = 7
number is 975
z = 6, x = 10, not possible
so the cases that work are:
531, 642, 753, 864, and 975
x , y, and z respectively .
so the original number is
100x + 10y + z
the number reversed is
100z + 10y + x
100x + 10y + z - (100z + 10y + x) = 396
99x - 99z = 396
x - z = 4
x = z+4
z - y = y - x
2y = z+x
y = (z+x)/2
" if the product of 3 digits number is non zero "
What does that mean??
xyz ≠ 0 ???
cases:
z = 0, x = 4 , y = 2
number is 420 , but the product is zero
z = 1 , x = 5, y = 3
number is 531
z = 2 , x = 6, y = 4
number is 642
z = 3 , x = 7, y = 5
number is 753
z = 4, x = 8 , y = 6
number is 864
z = 5, x = 9, y = 7
number is 975
z = 6, x = 10, not possible
so the cases that work are:
531, 642, 753, 864, and 975
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