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

Good! ^^

\(a,b\in\) { 1 ; - 1 ; 2 ; -2 ; -5 ; 5 ; 10 ; -10 ; -25 ; 25 ; 50 ; -50 } 

1 l i k e ! 

BCNN (a,b) = 50 => a,b = 25;50 . 10;50 . 1;50 . 2;50 . 5;50 

Good  

Because the greatest common dividor of m and n is 15, we put:

 {m=15kn=15h

(GCD(k;h)=1)

⇒3m+2n=45k+30h=225

⇒15(3k+2h)=225⇒3k+2h=15

+ If h = 0 then k = 5; and that result doesn't satisfy GCD(k;h) = 1.

So h > 0; then k is an odd number. 

3k<15⇒k<5⇒k∈{1;3}

If k = 1 then h = 6⇒

 m = 15 ; n = 90 ⇒mn=15.90=1350

If k = 3 then h = 3; that result doesn't satisfy GCD(k;h)=1.

Therefore, the answer is 1350.

Because the greatest common dividor of m and n is 15, we put:

 \(\left\{{}\begin{matrix}m=15k\\n=15h\end{matrix}\right.\)\(\left(GCD\left(k;h\right)=1\right)\)

\(\Rightarrow3m+2n=45k+30h=225\)

\(\Rightarrow15\left(3k+2h\right)=225\Rightarrow3k+2h=15\)

+ If h = 0 then k = 5; and that result doesn't satisfy GCD(k;h) = 1.

So h > 0; then k is an odd number. 

\(3k< 15\Rightarrow k< 5\Rightarrow k\in\left\{1;3\right\}\)

If k = 1 then h = 6\(\Rightarrow\) m = 15 ; n = 90 \(\Rightarrow mn=15.90=1350\)

If k = 3 then h = 3; that result doesn't satisfy GCD(k;h)=1.

Therefore, the answer is 1350.

We have : ƯCLN(140;240) = 20

Dress the largest square can be 20 cm x 20 cm

We have : ƯCLN(140;240) = 20

Dress the largest square can be 20 cm x 20 cm

We have : ƯCLN(140;240) = 20

Dress the largest square can be 20 cm x 20 cm

1339 = 103 x 13 = 1 x 1339.

No one can be 1339 years old, so the grandfather will be 103 years old and his grandchild will be 13 years old next year.

Thus, the grandfather is 102 years old, and his grandchild is 12 years old now.

1339 = 103 x 13 = 1 x 1339.

No one can be 1339 years old, so the grandfather will be 103 years old and his grandchild will be 13 years old next year.

Thus, the grandfather is 102 years old, and his grandchild is 12 years old now.

1339 = 103 x 13 = 1 x 1339.

No one can be 1339 years old, so the grandfather will be 103 years old and his grandchild will be 13 years old next year.

Thus, the grandfather is 102 years old, and his grandchild is 12 years old now.

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

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

Let 140a be the number of refrigerators Alvin sold in 2009, where a is a constant value. Thus,

- Ben sold 140a ÷

 7 = 20a (refrigerators);

- Carl sold 140a ÷

 5 = 28a (refrigerators);

- Dan sold 140a ÷

 4 = 35a (refrigerators).

According to the problem, we have,

140a + 20a + 28a + 35a = 669

223a = 669

a = 3.

Therefore, Alvin had sold at most 3 ×

 140 = 420 refrigerators.
Answer : 420

Let 140a be the number of refrigerators Alvin sold in 2009, where a is a constant value. Thus,

- Ben sold 140a \(\div\) 7 = 20a (refrigerators);

- Carl sold 140a \(\div\) 5 = 28a (refrigerators);

- Dan sold 140a \(\div\) 4 = 35a (refrigerators).

According to the problem, we have,

140a + 20a + 28a + 35a = 669

223a = 669

a = 3.

Therefore, Alvin had sold at most 3 \(\times\) 140 = 420 refrigerators.

Answer. 420 refrigerators

Let n be the smallest value of that number. When n is divided by 2,3,4,5,6,the remainder is always 1,so n - 1 is divisible by 2,3,4,5,6

=> n - 1 = LCM(2,3,4,5,6) = 60 => n = 61

Let n be the smallest value of that number. When n is divided by 2,3,4,5,6,the remainder is always 1,so n - 1 is divisible by 2,3,4,5,6

=> n - 1 = LCM(2,3,4,5,6) = 60 => n = 61

Let n be the smallest value of that number. When n is divided by 2,3,4,5,6,the remainder is always 1,so n - 1 is divisible by 2,3,4,5,6

=> n - 1 = LCM(2,3,4,5,6) = 60 => n = 61

We have : 882 = 2 x 32 x 72 ; 1134 = 2 x 34 x 7.So :

GCD(882 ; 1134) = 2 x 32 x 7 = 126

LCM(882 ; 1134) = 2 x 34 x 72 = 7938

We have : 882 = 2 x 3² x 7² ; 1134 = 2 x 3⁴ x 7.So :

GCD(882 ; 1134) = 2 x 3² x 7 = 126

LCM(882 ; 1134) = 2 x 3⁴ x 7² = 7938