2 x 2 + 1 = 5

3 x 2 + 1 = 7

5 x 2 + 1 = 11

7 x 2 + 1 = 15

11 x 2 + 1 = 23

13 x 2 + 1 = 27

So,there are 4 jillyprimes which are less than 15 : 2 ; 3 ; 5 ; 11 

2 x 2 + 1 = 5

3 x 2 + 1 = 7

5 x 2 + 1 = 11

7 x 2 + 1 = 15

11 x 2 + 1 = 23

13 x 2 + 1 = 27

So,there are 4 jillyprimes which are less than 15 : 2 ; 3 ; 5 ; 11

We have \(\overline{aabb}=11\times\overline{a0b}=11\times\left(100a+b\right)\)   

11 divide by 4 give we remainder 3. A square number divide by 4 give me remainder 0 or 1 

==> Look in 100a + b . \(100a⋮4\) so if b : 4 give we re. 1 or 2 the 11(100a + b ) gives re. 3 or 2 (no!)

So b : 4 must gives re. 3 or 4 , b is a digit ==> b = { 3 ; 4 ; 7 ; 8}. Square never have last dg. is 3;7 and 8 , so b =4

With sentences above we get 100a + b (100a + 4) divide by 3 gives re. 2 or 0. Remember b = 4 = 3 + 1, so a : 3 gives re. 1or 2. Grouping a we get a = {1 ; 2 ; 4; 5; 7; 8}. Try each a you'll find a=7

That number is 7744. A sq. of 88

We have : abcabc = abc x 1001 = abc x 7 x 11 x 13

Because abc is a prime number,the number of divisors of abcabc is :

(1 + 1)4 = 16

Because abc is a prime number,the number of divisors of abcabc is :

(1 + 1)4 = 16

We have : abcabc = abc x 1001 = abc x 7 x 11 x 13

Because abc is a prime number,the number of divisors of abcabc is :

(1 + 1)4 = 16

Your problem is not true, because 3210 = 2 x 3 x 5 x 107, and noone in this problem can be 107 years old!

\(300=2^2.3.5^2\)

The number of integer factors 300 have is \(2.\left[\left(2+1\right)\left(1+1\right)\left(2+1\right)\right]=36\)

Ans : 36.

You have : 300 = 22.3.52

The number of integer factors 300 have is :

                      2.[(2 + 1)(1 + 1)(2 + 1)] = 36

                               Answer : 36. good

We have : 300 = 22.3.52

The number of integer factors 300 have is :

                      2.[(2 + 1)(1 + 1)(2 + 1)] = 36

                               Answer : 36. good

We have : 37 = 3 + 5 + 29 (c1)

                 37 = 3 + 11 + 23 (c2)

                 37 = 5 + 13 + 19 (c3)

Drees is three ways are there to write 37

We have : 37 = 3 + 5 + 29 (c1)

                 37 = 3 + 11 + 23 (c2)

                 37 = 5 + 13 + 19 (c3)

Drees is three ways are there to write 37 

We have : 37 = 3 + 5 + 29 (c1)

                 37 = 3 + 11 + 23 (c2)

                 37 = 5 + 13 + 19 (c3)

Drees is three ways are there to write 37 

We have : 7m + n = mn + 11

=> 7m + n - mn = 11

=> m(7 - n) - 7 + n = 4

=> (m - 1)(7 - n) = 4

So we have a table as shown :

m - 1 1 2 4
7 - n 4 2 1
m 2 3 5
n 3 5 6 (absurd)
7m + n 17 26 (absurd) x

Hence,we have m = 2 ; n = 3 and m2 + n2 = 13

We have : 7m + n = mn + 11

=> 7m + n - mn = 11

=> m(7 - n) - 7 + n = 4

=> (m - 1)(7 - n) = 4

So we have a table as shown :

Hence,we have m = 2 ; n = 3 and m² + n² = 13

 

Only 1997 is a prime number

Only 1997 is a prime number