Let the ten's digit be x. Then, unit's digit = 2x + 1.
[10x + (2x + 1)] - [{10 (2x + 1) + x} - {10x + (2x + 1)}] = 1
<=> (12x + 1) - (9x + 9) = 1 <=> 3x = 9, x = 3.
So, ten's digit = 3 and unit's digit = 7. Hence, original number = 37.
We know that,
? Each digit has a fixed position called its place.
? Each digit has a value depending on its place called the place value of the digit.
? The face value of a digit for any place in the given number is the value of the digit itself
? Place value of a digit = (face value of the digit) × (value of the place).
Hence, the place value of 6 in 64 = 6 x 10 = 60.
Let the three integers be x, x + 2 and x + 4. Then, 3x = 2 (x + 4) + 3 <=> x = 11.
Third integer = x + 4 = 15.
Let the numbers be 4x, 5x and 6x, Then, (4x + 5x + 6x ) / 3 = 25
=> 5x = 25
=> x = 5.
Largest number 6x = 30.
Let the numbers be x and 1365+x
Then 1365+x = 6x+15
=> x = 270
Let the numbers be x and y.
Then, xy = 9375 and x/y = 15.
=>
=> y = 25
=> x = 15y = 15 x 25 = 375.
Sum of the numbers = 375 + 25 = 400.
Let the ten's digit be x.
Then, unit's digit = x + 2.
Number = 10x + (x + 2) = 11x + 2
Sum of digits = x + (x + 2) = 2x + 2.
(11x + 2) (2x + 2) = 144
=> 22x^2+26x-140=0
=> (x - 2)(11x + 35) = 0
=> x = 2
Hence, Required Number = 11x + 2 = 24
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