Let the smaller number be 'm'
Then, from the given data, the larger number is '3m'
Given that m + 3m = 44
=> 4m = 44
m = 44/4 = 11
=> m = 11
=> 3m = 3 x 11 = 33
Hence, the two numbers are 11 and 33.
(300 ? 101)/11 = 199/11 = 18 1/11
18 Numbers.
By hit and trail, we find that
47619 x 7 = 333333.
7) 333333 (47619
333333
----------------
0
And 476190476 x 7 = 3333333333 but smallest number is 47619.
Let the greater and smaller number be p and q respectively.
4q = p + 5 ------ (I)
p = 3q+2 ------- (II)
From equation (I) and (II)
q = 7
p = 23
Let the numbers be x and y. Then,
x^2+y^2=3341......(1)
x^2-y^2=891......(2)
Adding (i) and (ii), we get : 2x^2=4232 or x^2= 2116 or x =46
Subtracting (ii) from (i), we get : 2y^2= 2450 or y^2 = 1225 or y = 35
So, the numbers are 35 and 46
Let the number be x. Then, 15x - x = 196 <=> 14x = 196 => x = 14..
Multiplying two negatives (or any even multiple) results in a positive. But multiplying three negatives (or any odd multiple) gives a negative. If the result of multiplying 6 negatives is odd, the largest number of negative integers will be the largest odd number (i.e.5)
Let the two no's be a and b;
Given product of the no's is p = ab;
If the each nos is increased by 2 then the new product will be
(a+2)(b+2) = ab + 2a + 2b + 4
= ab + 2(a+b) + 4
= p + 2(a+b) + 4
Hence the new product is (p+4) times greater than twice the sum of the two original numbers.
If a number to be divisile by 88, it should be divisible by both "8" and "11"
Check for '8' :
For a number to be divisible by "8", the last 3-digit should be divisible by "8"
Here 72x23y --> last 3-digit is '23y'
So y=2 [ (i.e) 232 is absolutely divisible by "8"]
Chech for '11' :
For a number to be divisible by "11" , sum of odd digits - sum of even digits should be divisible by "11"
(7 + x + 3) - (2 + 2 + y)
(7 + x + 3) - (2 + 2 + 2)
(10 + x) - 6 should be divisible by "11"
for x = 7
=> 17 - 6 = 11 [ which is absolutely divisible by "11"]
So x = 7 , y= 2.
Lowest 4-digit number is 1000.
LCM of 3, 4 and 5 is 60.
Dividing 1000 by 60, we get the remainder 40. Thus, the lowest 4-digit number that exactly divisible by 3, 4 and 5 is 1000 + (60 - 40) = 1020.
Now, add the remainder 2 that's required. Thus, the answer is 1022.
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