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Updated on: 03 Apr 2021, 01:51
Originally posted by MuktaSawant on 03 Apr 2021, 01:49.
Last edited by Bunuel on 03 Apr 2021, 01:51, edited 1 time in total.
Last edited by Bunuel on 03 Apr 2021, 01:51, edited 1 time in total.
Renamed the topic.
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When a two digit number is reversed, resulting number is another 2 digit number. what is the maximum difference between their squares?
A. 4860
B. 5480
C. 6440
D. 7920
E. 8460
A. 4860
B. 5480
C. 6440
D. 7920
E. 8460
03 Apr 2021, 08:38
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Given: When a two digit number is reversed, resulting number is another 2 digit number.
Asked: what is the maximum difference between their squares?
Let the 2 digit number be xy where x is tens digit and y is unit digit
Original number xy:
10x + y
Number with digits reversed yx:
10y + x
Difference between their squares
(10x+y)^2 - (10y+x)^2 = (11x+11y)(9x-9y) = 11*9(x+y)(x-y) = 99(x^2 - y^2)
x^2 - y^2 is maximum when x=9 & y=1
Maximum difference between their squares = 99(9^2 - 1^2) = 99*80 = 7920
IMO D
Asked: what is the maximum difference between their squares?
Let the 2 digit number be xy where x is tens digit and y is unit digit
Original number xy:
10x + y
Number with digits reversed yx:
10y + x
Difference between their squares
(10x+y)^2 - (10y+x)^2 = (11x+11y)(9x-9y) = 11*9(x+y)(x-y) = 99(x^2 - y^2)
x^2 - y^2 is maximum when x=9 & y=1
Maximum difference between their squares = 99(9^2 - 1^2) = 99*80 = 7920
IMO D
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03 Apr 2021, 10:41
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let the number be xy= 10x+y
reverse = yx= 10y+x
now if we look at difference
the diff = 9(x-y)
x-y will be max when x= 9 and y = 1
so number is 91 and 19
\(91^2 \)= 8281
\(19^2\) = 361
thus diff = 7920
D IMO
reverse = yx= 10y+x
now if we look at difference
the diff = 9(x-y)
x-y will be max when x= 9 and y = 1
so number is 91 and 19
\(91^2 \)= 8281
\(19^2\) = 361
thus diff = 7920
D IMO
