Then, xy = 120 and x2 + y2 = 289.
∴ (x + y)2 = x2 + y2 + 2xy = 289 + (2 x 120) = 529
∴ x + y = √529 = 23.
Let ten's and unit's digits be 2x and x respectively.
Then, (10 x 2x + x) - (10x + 2x) = 36
⟹ 9x = 36
⟹ x = 4.
∴ Required difference = (2x + x) - (2x - x) = 2x = 8.
Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131.
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400.
⟹ (a + b + c) = √400 = 20.
Then, x + 17 = | 60 |
x |
⟹ x2 + 17x - 60 = 0
⟹ (x + 20)(x - 3) = 0
⟹ x = 3.
Then, (x + 2)2 - x2 = 84
⟹ 4x + 4 = 84
⟹ 4x = 80
⟹ x = 20.
∴ The required sum = x + (x + 2) = 2x + 2 = 42.
To find the largest 4 digit number exactly divisible by 88,
We should divide the largest possible 4 digit number by 88, and if we get any remainder than subtract it from that largest number.
The largest possible 4 digit number is 9999
Now,
88) 9999 (113
88
_______
119
88
_______
319
264
_______
55
_______
Therefore, the largest 4 digit number exactly divisible by 88 is given by
9999 - 55 = 9944.
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 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.
Let the three integers be x, x + 2 and x + 4. Then, 3x = 2 (x + 4) + 3 <=> x = 11.
Third integer = x + 4 = 15.
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