percom and numbers

2²â‰¡0mod4
2^2k≡0mod4
5≡1mod4
5^l≡1mod4
3^m≡(-1)^mmod4
hence for all
p (except 0 and 1)
r→ Z⁺
m=odd

now case two
p=1
2≡2mod4
m=even
l→Z+
its again divisible by 4

No , the answer is 500.

small question -- WHY ISN'T ANYBODY TRYING ????????????

Seems I only have to give the answer ------

( the sign ‘ iotf ’ represents “ is of the form “ )

3^odd iotf 4n + 3 , 5^any iotf 4p + 1 , 3^even iotf 8m + 1 ,

Case : 1 . When p = 1 ,

2^p + 5^r + 3^q iotf 2 + 4p + 1 + 3^q iotf 4p + 3 + 3^q .

Obviously , to make 2^p + 3^q + 5^r a multiple of 4 , we have to have

3^q of the form 8m + 1 , because then

2^p + 5^r + 3^q iotf 4p + 3 + 8m + 1 iotf 4p + 8m + 4 iotf 4a .

So q should be even , hence it can have the values 2 , 4 , 6 , 8 , 10 , i.e , it can take up 5

values .

r can take any value in the range 1-10 , hence it can have 10 values .

p here is a fixed number , hence it takes only 1 value .

So total number of such numbers in case 1 …. 10 x 5 x 1 = 50 .

Case : 2 > When 2 <= p <= 10 , 2^p iotf 4x .

Hence 2^p + 5^r + 3^q iotf 4x + 4p + 1 + 3^q ,

So here 3^q should be of the form 4n + 3 .

Hence q has to be odd , so it can take 5 values .

As 2<= p <= 10 , so p can take 9 values .

r can take any of the arbitrary 10 values .

Hence number of such numbers in case 2 : > 10 x 9 x 5 = 450

Hence total number of such numbers

= number of such numbers obtained in ( case 1 + case 2 )

= 450 + 50

= 500