Assume sum of digits as x+y. It is given x+y=10. Assume those two digit no as 10x+y reverse of it is 10y+x
It is given if you reverse it new no will be 54 less than original no
So
10y+x=10x+y-54
-9x+9y=-54
-9(x-y)=-54
x-y=6
Equate x+y and x-y
x+y-10=x-y-6
x+y-10-x+y+6=0
2y-4=0
y=2 substitute this in x+y =10
x+2=10
x=8
Substitute this x and y in two digit no we assumed that is 10x+y
10*8+2= 82 is original no
Reverse of it 28
Ans is 82
82
8+2=10
82-28=54
192
Possibility of two digit no whose sum is 17 89 and 98 Reverse of 89 is 98. 98 is 9 less than the original no 89. 89 is original no
11yrs.
Find a four digit number whose digits will be reversed when multiplied by nine?
Possibility of two digit no whose sum is 10 19,28,37,46,55,64,73,82,91 Now add 54 to each no mentioned above 73,82,91,100,109,118,127,136,145 See after 1st comma 28 and 82 Reverse of 28 is 82. That no 82 is 54 more than the no 28. So 28 is the original
63
The number is 36
17
192
If the number with the digits reversed can have a leading 0 so that it is a 1-digit number, then 16. Otherwise 13.
five digits number when you multiply by four is the same the number when you reversed it is 21978*4 = 87912
Possibility of two digit no whose sum is 17 89 and 98 Reverse of 89 is 98. 98 is 9 less than the original no 89. 89 is original no
An eight digit number with one zero cannot remain the same when its digits are reversed. It must have an even number of 0s.
after 11yrs.
11yrs.
47 Impossible problem!
Find a four digit number whose digits will be reversed when multiplied by nine?