What is the least common multiple of 152 and 190? A. 304 : PS Archive
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# What is the least common multiple of 152 and 190? A. 304

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What is the least common multiple of 152 and 190? A. 304  [#permalink]

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31 Oct 2005, 14:22
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What is the least common multiple of 152 and 190?

A. 304
B. 14440
C. 760
D. 28880
E. 1520
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31 Oct 2005, 14:29
152 = 2 x 2 x 2 x 19
190 = 2 x 5 x 19

LCM = 2 x 2 x 2 x 5 x 19 --> <b>C. 760</b>
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31 Oct 2005, 15:28
This is a good one!
Good work ValleyBall1

I incorrectly picked B. 19 the prime number is the key here but I realized it's difficult to chase such primes. Instead attack the A/C's starting with the least number and divide each with 152,190 and see which one cuts even.
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31 Oct 2005, 15:58
Good point mbaqst! I didn't think about that. Since 152 is even, I decided to start with 2 and go from there. Eventually I came up with my prime box. Since 190 ends in 0, I started with 10 and went from there.

Memorizing the divisibility number properties helps...

<b>Divisible by 2</b> if the number is even

<b>Divisible by 3</b> if the sum of the digits is divisible by 3

<b>Divisible by 4</b> if the number can be cut in half twice

<b>Divisible by 5</b> if the number ends with 0 or 5

<b>Divisible by 6</b> if the number is divisible by 2 & 3

<b>Divisible by 8</b> if the number can be cut in half three times

<b>Divisible by 9</b> if the sum of the digits is divisible by 9

Other than multiples of 7, I'm not aware of any other rules for 7. Anyone have shortcuts for 7?
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31 Oct 2005, 19:03
Good work, OA is C.

Here is the detailed explanation:
The least common multiple (LCM) of two numbers is the smallest number that the two numbers will divide into evenly.

To find the LCM of 2 numbers:

1. Find the greatest common factor (GCF) of the numbers
2. Multiply the numbers together
3. Divide step 2 by step 1

Applying these rules to this problem:

1. Find the greatest common factor (GCF) of the numbers

To find the greatest common factor:

A) Find the prime factorization of each number
B) Multiply the factors that both numbers have in common

Prime factorization of 152: 2 * 2 * 2 * 19
Prime factorization of 190: 2 * 5 * 19

Multiplying the factors in common: 2 * 19 = 38

2. Multiply the numbers together

152 * 190 = 28880

3. Divide step 2 by step 1

28880 / 38 = 760
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31 Oct 2005, 21:43
ValleyBall1 wrote:
<b>Divisible by 4</b> if the number can be cut in half twice

<b>Divisible by 4</b> if the number can be cut in half twice

<b>Divisible by 8</b> if the number can be cut in half three times

Other tricks:
Divisible by 4 if the last two digits of the number for a number which is divisible by 4
for example : 116 , 16 is divisible by 4 ---> 116 is divisible by 4.
Divisible by 8 if the last three digits of the number form a number which is divisible by 8for example: 1235745354543256 , 256 is divisible by 8 ---> 1235745354543256 is divisible by 8.
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31 Oct 2005, 21:52
ValleyBall1 wrote:
Good point mbaqst! I didn't think about that. Since 152 is even, I decided to start with 2 and go from there. Eventually I came up with my prime box. Since 190 ends in 0, I started with 10 and went from there.

Memorizing the divisibility number properties helps...

<b>Divisible by 2</b> if the number is even

<b>Divisible by 3</b> if the sum of the digits is divisible by 3

<b>Divisible by 4</b> if the number can be cut in half twice

<b>Divisible by 5</b> if the number ends with 0 or 5

<b>Divisible by 6</b> if the number is divisible by 2 & 3

<b>Divisible by 8</b> if the number can be cut in half three times

<b>Divisible by 9</b> if the sum of the digits is divisible by 9

Other than multiples of 7, I'm not aware of any other rules for 7. Anyone have shortcuts for 7?

To check if a number is divisible by 7, multiple the last digit by 2 and subtract it from the rest of the number. If the remaining number is divisible by 7, the number is too.

For example:

Is 3675 divisible by 7?

(1) Multiply 5 with 2. We get 10
(2) Subtract 10 from 367. We get 357.
(3) Is 357 divisible by 7? Yes, we get 51.
(4) So 3675 is divisible by 7. (3675/7 = 525)
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31 Oct 2005, 22:44
Good one Valleyball! Your diagramming is exactly the same as the OE. I "backsolved" this one by starting with C and got lucky.
31 Oct 2005, 22:44
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