Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = ( 22 + x ), which must be divisible by 3.
When x=2. [(22 + 2) = 24 is divisible by 3]
So, the answer is 2
132 = 4 x 3 x 11, So if the number is divisible by all three numbers 4,3 and 11,then the number is divisible by 132 also.
264 => 4,3,11(/)
396 => 4,3,11(/)
462 => 11,3
792 => 4,3,11(/)
968 => 11,4
2178 => 11,3
5184 => 3,4
6336 => 4,3,11(/)
Required number of numbers=4.
6)4456(742
42
--------
25
24
-------
16
12
-----
4
------
Required number = (6-4) = 2.
Let a and b are the numbers.Then a+b is 12 and ab is 35.
a+b/ab = 12/35
1/b + 1/a = 12/35
LCM of 2, 3, 7 is 42.
=> (700 ? 300)/42 = 9 22/42 => 9 Numbers.
The largest 4 digit number is = 9999
After dividing 9999 with 88, we get 55 as remainder
so largest 4 digit number divisible by 88 = 9999-55 = 9944.
The maximum number of bows will be 4 yards (= 4 x 36 inches) divided by 15 inches.
This gives 9.6. But as a fraction of a bow is no use, we can only make 9 bows.
Given that,
Now Power Cycle of
4 is 4, 6
6 is 6
8 is 8, 4, 2, 6
1 is 1
- 198 is even number , So the unit digit value is 6
- It is always 6
- 66/4 the remainder is 2 , So the second value is 4
- 11 is odd number, So the unit digit value is 4
1 - It is always 1
Total = 21
Therefore, The unit digit is 1
Let n=4*q + 3. Then, 2*n = 8*q + 6 = 4(2*q + 1) + 2.
Thus when 2*n is divided by 4, the reminder is 2.
Let,
no. of rice bowl = x
no. of dal bowl = y
no. of meat bowl = z
Then, x + y + z = 65 ....(1)
Also given that, 2x = 3y = 4z
substituting value of y & z in terms of x i.e. y = 2x/3 and z = x/2 in eq (1),
We get x = 30.
As 1 rice bowl is shared between 2 guests,
Therefore, there are 60 guests.
On dividing we get
75)8485(113
75
------
98
75
------
235
225
------
10
------
Required number = (8485-10)=8475.
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