Prime Numbers and Composite Numbers
Đỗ Phi Phi
26/07/2017 at 21:39
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
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FA KAKALOTS 28/01/2018 at 22:04
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
steve jobs
27/03/2017 at 11:05steve jobs
27/03/2017 at 10:56
Nguyen Trong Quang 05/09/2019 at 15:40
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
steve jobs
27/03/2017 at 10:56steve jobs
27/03/2017 at 10:54steve jobs
27/03/2017 at 10:53steve jobs
27/03/2017 at 10:52
»ﻲ2004#ﻲ« 29/03/2017 at 06:10
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

Number One 02/04/2017 at 06:29We 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

Phạm Tuấn Kiệt 18/05/2017 at 19:49
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
steve jobs
27/03/2017 at 10:51
Nguyễn Nhật Minh 01/06/2017 at 12:23
Your problem is not true, because 3210 = 2 x 3 x 5 x 107, and noone in this problem can be 107 years old!
Bill gates
20/03/2017 at 10:51Bill gates
20/03/2017 at 10:49
Run my EDM 20/03/2017 at 12:40
\(300=2^2.3.5^2\)
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.
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→இے๖ۣۜQuỳnh 22/03/2017 at 20:18
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

»ﻲ†hïếu๖ۣۜGïลﻲ« 25/03/2017 at 18:59
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
Bill gates
20/03/2017 at 10:49
¤« 24/02/2018 at 13:57
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

Lê Nho Khoa 23/03/2017 at 21:07
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

Nguyễn Việt Hoàng 21/03/2017 at 06:12
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
Bill gates
20/03/2017 at 10:48
¤« 24/02/2018 at 13:57
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) xHence,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 :
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 m^{2} + n^{2} = 13
Bill gates
20/03/2017 at 10:46
¤« 24/02/2018 at 13:57
Only 1997 is a prime number

Only 1997 is a prime number