Numbers problems

Last digit for the power of 6 is 6 (always)

Power cycle of 7 is 7, 9, 3, 1.

Now 467/4 gives a remainder of 3

Then the last digit is 7 3 = 3

Last digit is 6 + 3 = 9.

Screws weight = 90x100 = 9000gms
bolts weight = 100 x 150 = 15000gms
weight = screws + bolts => 24000 gms =>24 kg
Given entire box weight = 35.5kg
empty box = entire box weight - weight => 35.5kg - 24kg => 11.5kg
so the empty box is 11.5kg.

1 to 300 = 300 numbers
squares btwn 1 to 300 = 1,4,9,16,25,36,49.....289 i.e, total 17
cubes = 1,8,27,64,125,216 i.e, 5
now 1 and 64 are common so total numbers that should be removed ( 17+5-2= 20 )
total 20 term removed so 300th term would be 320

A prime number is a whole number greater than 1 whose only factors are 1 and itself.

Factors of 5 are 1, 5

Factors of 11 are 1, 11

Factors of 21 are 1, 3, 7, 21

Factors of 37 are 1, 37.

Hence, according to the definition of a prime number, 21 is not a prime number as it has more than two factors.

   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.

Given   6 5 × 35 9 × 11 3

? 2 × 3 5 × 5 × 7 9 × 11 3 

? 2 5 × 3 5 × 5 9 × 7 9 × 11 3

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

Let
x = peacocks
y = deers

2x + 4y = 200
x + y = 72
y = 72 - x
2x + 288 - 4x = 200
x = 44 peacocks
y = 28 deers

87)13601(156
       87
       -----
       490
       435
        -----
        551
        522
         -----
         29
         -----
Required number is 29.

Required numbers are 10,15,20,25,...,95
This is an A.P. in which a=10,d=5 and l=95.
Let the number of terms in it be n.Then t=95
So a+(n-1)d=95.
10+(n-1)*5=95,then n=18.
Required sum=n/2(a+l)=18/2(10+95)=945.

(Place value of 7)-(face value of 7)
=7000-7=6993.