Find the next number in the series 125, 80, 45, 20, ....

 

The pattern is three consecutive numbers and then three consecutive numbers missing.Then four consecutive numbers and then four consecutive numbers missing.Then five consecutive numbers and then five consecutive numbers missing.

The next number=20+6=26

When we divide 1234 by 8, remainder is 2

When we divide 1235 by 8, remainder is 3

When we divide 1237 by 8, remainder is 5

---> 2 x 3 x 5 = 30 

As 30 will not be the remainder because it is greater than 8,

when 30 divided by 8, remainder = 6.

Let x be the number.

Given: x = 296n + 75 [where n is quotient] 

 --> (378n) +( 372 + 1) 

 -->37(8n + 2) + 1

Hence, when x divided by 37, 1 is the remainder.

1523+2323=19-423+19+423  

 The remainder on dividing by 19 will be -423+423= 0.

When 16nis divided by 9, we have

 1619, remainder = 7  

1629, remainder = 4 

1639, remainder = 1 

1649, remainder = 7 

1659, remainder = 4 

1669, remainder = 1 

So, we have cyclicity of 3 factors i.e 7,4,1. 

Hence only 3 remainders are possible.

Given that, 4198+612345+34866+2411+1

Now Power Cycle of

4 is 4, 6 

6 is 6 

8 is 8, 4, 2, 6

1 is 1

4198- 198 is even number , So the unit digit value is 6 

612345 - It is always 6

34866-  66/4 the remainder is 2 , So the second value is 4

2411- 11 is odd number, So the unit digit value is 4

1 - It is always 1 

Total = 21

Therefore, The unit digit is 1

Given  65×359×113

⇒2×35×5×79×113 

⇒25×35×59×79×113

Thus, there are (5 + 5 + 9 + 9 + 3)= 31 prime numbers.

   1 1
   8 A 9
+ 4 C 6
-  6 B 2
   7 2 3
We may represent the given sum, as shown below:

1 + A + C - B = 12
A + C - B = 11

Now,giving the maximum values to A and C, i.e.

A = 9 and C = 9, we get B = 7.