Question 137560
I rewrote both problems slightly.
:
The sum of the digits is 8. If 18 is added to the number , the digits of the new numeral are those of the original numeral, but they appear in reversed order.
:
Let x = 10's digit of the original number
Let y = units digit of the original number
The number = 10x + y
:
It says the sum of the digits is 8:
x + y = 8
or
y = (8-x); use this for substitution
;
"If 18 is added to the number , the digits of the new numeral are those of the original numeral, but they appear in reversed order."
10x + y + 18 = 10y + x
10x - x + 18 = 10y - y
9x + 18 = 9y
Simplify, divide equation by 9
x + 2 = y
Substitute (8-x) for y
x + 2 = 8 - x
x + x = 8 - 2
2x = 6
x = {{{6/2}}}
x = 3
Then y = 5; the original number = 35
:
Check solution in the statement;
"If 18 is added to the number , the digits of the new numeral are those of the original numeral, but they appear in reversed order."
18 + 35 = 53; confirms our solution
:
:
The sum of the digits is 8. The number is 16 less than the product of 3 and the number represented when the digits are interchanged.
:
Let x = 10's digit of the original number
Let y = units digit of the original number
The number = 10x + y
:
It says the sum of the digits is 8:
x + y = 8
or
y = (8-x); use this for substitution
;
"The number is 16 less than the product of 3 and the number represented when the digits are interchange"
10x + y = 3(10y+x) - 16
10x + y = 30y + 3x - 16
10x - 3x = 30y - y - 16
7x = 29y - 16
Substitute (8-x) for y:
7x = 29(8-x) - 16
7x = 232 - 29x - 16
7x + 29x = 232 - 16
36x = 216
x = {{{216/36}}}
x = 6
then y = 2; the number = 62
:
Check solution in the statement:
"The number is 16 less than the product of 3 and the number represented when the digits are interchange"
62 = 3(26) - 16
62 = 78 - 16; confirms out solutions