GCD and LCM

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this question very easy

There are 32 segments . 

please, me off.

Because the lowest common multiple of a and b is 50 . So all possible values of a and b is :

(a;b) = {1 ; 2 ; 5 ; 10 ; 25 ; 50}

Because the lowest common multiple of a and b is 50 . So all possible values of a and b is :

(a;b) = {1 ; 2 ; 5 ; 10 ; 25 ; 50}

Because the lowest common multiple of a and b is 50 . So all possible values of a and b is :

(a;b) = {1 ; 2 ; 5 ; 10 ; 25 ; 50}

m=45;n=45        

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Let a (cm) be the largest possible square size

=> a = GCD (140 ; 240) = 20 

So the largest size of the square's edge is 20 cm.Then the smallest number of squares can be cut from the cardboard is :

140 x 240 : 20² = 84

Answer : 20 cm ; 84 squares

Hello,my friend

Let a (cm) be the largest possible square size

=> a = GCD (140 ; 240) = 20 

So the largest size of the square's edge is 20 cm.Then the smallest number of squares can be cut from the cardboard is :

140 x 240 : 202 = 84

Answer : 20 cm ; 84 squares

1330 = 2 x 5 x 7 x 19 = 2 x 7 x 5 x 19 = 14 x 95

So the grandpa's age and his grandchild's age next year should be 95 and 14 respectively.Then now,the grandpa is 94 and his grandchild is 13

1998 = 2 x 3³ x 37,then 1998 has : (1 + 1)(3 + 1)(1 + 1) = 16 (factors)

So x = 16

P/S : 16 is the number of positive factors of 1998,so 1998 has 32 integer factors

1998 = 2 x 33 x 37,then 1998 has : (1 + 1)(3 + 1)(1 + 1) = 16 (factors)

So x = 16

P/S : 16 is the number of positive factors of 1998,so 1998 has 32 integer factors

1998 = 2 x 33 x 37,then 1998 has : (1 + 1)(3 + 1)(1 + 1) = 16 (factors)

So x = 16

P/S : 16 is the number of positive factors of 1998,so 1998 has 32 integer factors

3432 = 2³ x 3 x 11 x 13

5460 = 2² x 3 x 5 x 7 x 13

=> GCD (3432 ; 5460) = 2² x 3 x 13 = 156

5460 = 22 x 3 x 5 x 7 x 13

=> GCD (3432 ; 5460) = 22 x 3 x 13 = 156

2000=2^4 * 5^3 = 2*2*2*2*125 = 5*5*5*2*8

=> min (a+b+c+d)=5+5+5+2+8=25

2000=2^4 * 5^3 = 2*2*2*2*125 = 5*5*5*2*8

=> min (a+b+c+d)=5+5+5+2+8=25

To use the smallest number of tiles,the size of each tile must be the largest.Let x (cm) be that size,so we have x = GCD(30 ; 45) = 15

=> y = 30 x 45 : 15² = 6

To use the smallest number of tiles,the size of each tile must be the largest.Let x (cm) be that size,so we have x = GCD(30 ; 45) = 15

=> y = 30 x 45 : 152 = 6

To use the smallest number of tiles,the size of each tile must be the largest.Let x (cm) be that size,so we have x = GCD(30 ; 45) = 15

=> y = 30 x 45 : 152 = 6

Let [a,b] be the LCM of a and b, and (a,b) be the GCD of a and b.

Since (a,b) = 8, [a,b] = 96, then a ×

 b = [a,b] × (a,b) = 96 ×

 8 = 768.

As (a,b) = 8, let a = 8m, b = 8n, in which (m,n) = 1.

=> a ×

 b = 8m × 8n = 768. Thus, m × n = 768 ÷ (8 ×

 8) = 12.

(m,n) = 1, so all the pairs (m;n) which have GCD = 1 and m ×

 n = 12 are:

(1;12),(3;4),(4;3),(12;1).

=> (a;b) = (8;96),(24;32),(32;24),(96;8).

Therefore, there are 2 possible values of a + b: 8+ 96 = 104, 24 + 32 = 56.

Answer. 104 and 56

Let [a,b] be the LCM of a and b, and (a,b) be the GCD of a and b.

Since (a,b) = 8, [a,b] = 96, then a \(\times\) b = [a,b] \(\times\) (a,b) = 96 \(\times\) 8 = 768.

As (a,b) = 8, let a = 8m, b = 8n, in which (m,n) = 1.

=> a \(\times\) b = 8m \(\times\) 8n = 768. Thus, m \(\times\) n = 768 \(\div\) (8 \(\times\) 8) = 12.

(m,n) = 1, so all the pairs (m;n) which have GCD = 1 and m \(\times\) n = 12 are:

(1;12),(3;4),(4;3),(12;1).

=> (a;b) = (8;96),(24;32),(32;24),(96;8).

Therefore, there are 2 possible values of a + b: 8+ 96 = 104, 24 + 32 = 56.

Answer. 104 and 56

We have : 3 - 2 = 4 - 3 = 5 - 4 = 6 - 5 = 1

So n + 1 is divisible by 3,4,5,6 but we need to find the smallest value of n.

=> n + 1 = LCM(3,4,5,6) = 60 => n = 59

We have : 3 - 2 = 4 - 3 = 5 - 4 = 6 - 5 = 1

So n + 1 is divisible by 3,4,5,6 but we need to find the smallest value of n.

=> n + 1 = LCM(3,4,5,6) = 60 => n = 59