Sum of the digit of a number of two digit is 4.If the digit of the unit is removed from the tenth digit then 2 remains.Then find the number?
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Hlw mate!!
Let us assume the digits of a number as 'x' and 'y'. Since it is a 2 digit number, we have one's place and ten's place. ( i.e., The place value of one digit is multiple of 10 and the place value of other digit is one.) I assume you know the difference between place value and face value.
Now going to the problem,
Sum of two digits is 8. => x +y=8→A
Diff between number and its reverse:
=> (10x+y)-(10y+x)=18
=>9x-9y=18
=> x-y=(18/9)=2→B
Solving A and B
x+y=8
x-y=2
=>2x=10=>x=5. Hence y = 3.
Hence the required number is 53
Hope it helpful
Let us assume the digits of a number as 'x' and 'y'. Since it is a 2 digit number, we have one's place and ten's place. ( i.e., The place value of one digit is multiple of 10 and the place value of other digit is one.) I assume you know the difference between place value and face value.
Now going to the problem,
Sum of two digits is 8. => x +y=8→A
Diff between number and its reverse:
=> (10x+y)-(10y+x)=18
=>9x-9y=18
=> x-y=(18/9)=2→B
Solving A and B
x+y=8
x-y=2
=>2x=10=>x=5. Hence y = 3.
Hence the required number is 53
Hope it helpful
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