the sum of a two digit number and the number obtained by reversing the digits is 66. if the digits of the number differ by 2, find the number. how many such numbers are there?
Let the two-digit number be 10x+y10x+y10x+y, where xxx is the tens digit and yyy is the units digit. We are given:
- The sum of the number and its reverse is 66.
- The digits differ by 2.
Step 1: Form the equations
Original number: 10x+y10x+y10x+y
Reversed number: 10y+x10y+x10y+x Sum condition:
(10x+y)+(10y+x)=66⇒11x+11y=66⇒x+y=6(10x+y)+(10y+x)=66\Rightarrow 11x+11y=66\Rightarrow x+y=6(10x+y)+(10y+x)=66⇒11x+11y=66⇒x+y=6
Difference condition:
∣x−y∣=2|x-y|=2∣x−y∣=2
So we have two possible systems:
- x−y=2x-y=2x−y=2 and x+y=6x+y=6x+y=6
- y−x=2y-x=2y−x=2 and x+y=6x+y=6x+y=6
Step 2: Solve the systems
- x−y=2x-y=2x−y=2, x+y=6x+y=6x+y=6
Add them:
2x=8⇒x=42x=8\Rightarrow x=42x=8⇒x=4
Then 4+y=6⇒y=24+y=6\Rightarrow y=24+y=6⇒y=2
Number: 10x+y=4210x+y=4210x+y=42
- y−x=2y-x=2y−x=2, x+y=6x+y=6x+y=6
Add them:
2y=8⇒y=42y=8\Rightarrow y=42y=8⇒y=4
Then x+4=6⇒x=2x+4=6\Rightarrow x=2x+4=6⇒x=2
Number: 10x+y=2410x+y=2410x+y=24
Both satisfy the conditions:
- 42+24=6642+24=6642+24=66 and ∣4−2∣=2|4-2|=2∣4−2∣=2
- 24+42=6624+42=6624+42=66 and ∣2−4∣=2|2-4|=2∣2−4∣=2
Final answer
- The two-digit numbers are 42 and 24.
- There are 2 such numbers.
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