So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.
264 → 11,3,4 (/)
396 → 11,3,4 (/)
462 → 11,3 (X)
792 → 11,3,4 (/)
968 → 11,4 (X)
2178 → 11,3 (X)
5184 → 3,4 (X)
6336 → 11,3,4 (/)
Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.
Required number of number = 4.
∴ x = 3, as 632 is divisible 8.
Thus, when 2n is divided by 4, the remainder is 2.
Let the two consecutive odd integers be (2x + 1) and (2x + 3)
Then,
(2x + 3)2 - (2x + 1)2
= (2x + 3 + 2x + 1) (2x + 3 - 2x - 1)
= (4x + 4)(2)
= 8 (x + 1), which is always divisible by 8
The pattern is -45, -35, -25, -15
The next number = 20-15= 5
The pattern is ,
The next number= = 4964
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.
LCM of 2, 3, 7 is 42.
=> (700 ? 300)/42 = 9 22/42 => 9 Numbers.
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,
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.
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