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How many positive two-digit integers have a remainder of 1 w

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How many positive two-digit integers have a remainder of 1 w

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[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, and a remainder of 4 when divided by 5?

A. 1
B. 2
C. 3
D. 4
E. 5

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A quick lesson on remainders:

Quote:
When x is divided by 5, the remainder is 3.
In other words, x is 3 more than a multiple of 5:
x = 5a + 3.

When x is divided by 7, the remainder is 4.
In other words, x is 4 more than a multiple of 7:
x = 7b + 4.

Combined, the statements above imply that when x is divided by 35 -- the LOWEST COMMON MULTIPLE OF 5 AND 7 -- there will be a constant remainder R.
Put another way, x is R more than a multiple of 35:
x = 35c + R.

To determine the value of R:
Make a list of values that satisfy the first statement:
When x is divided by 5, the remainder is 3.
x = 5a + 3 = 3, 8, 13, 18...
Make a list of values that satisfy the second statement:
When x is divided by 7, the remainder is 4.
x = 7b + 4 = 4, 11, 18...
The value of R is the SMALLEST VALUE COMMON TO BOTH LISTS:
R = 18.

Putting it all together:
x = 35c + 18.

Another example:
When x is divided by 3, the remainder is 1.
x = 3a + 1 = 1, 4, 7, 10, 13...
When x is divided by 11, the remainder is 2.
x = 11b + 2 = 2, 13...

Thus, when x is divided by 33 -- the LCM of 3 and 11 -- the remainder will be 13 (the smallest value common to both lists).
x = 33c + 13 = 13, 46, 79...
Onto the problem at hand:

Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, and a remainder of 4 when divided by 5?

A. 1
B. 2
C. 3
D. 4
E. 5
a remainder of 1 when divided by 2
x = 2a + 1 = 1, 3, 5, 7...
In other words, x must be ODD.

a remainder of 4 when divided by 5
x = 5b + 4 = 4, 9, 14, 19...
Since x must be odd, we get:
x = 9, 19, 29....

a remainder of 2 when divided by 3
In the blue list above, 29 is smallest value that yields a remainder of 2 when divided by 3.

Thus, when x is divided by 30 -- the LCM of the three divisors 2, 3, and 5 -- the remainder will be 29 (the smallest value that satisfies all of the given conditions):
x = 30c + 29 = 29, 59, 89, 119...

In the resulting list of options for x, only the 3 values in green are two-digit integers.

The correct answer is C.

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Junior | Next Rank: 30 Posts Default Avatar
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the answer should be C.
Since we can break it down to 3ok +29.

Hence we get three 2 digit numbers

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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, and a remainder of 4 when divided by 5?

A. 1
B. 2
C. 3
D. 4
E. 5
$$?\,\,:\,\,\# \,\,N\,\,{\rm{with}}\,\,\left( 1 \right),\left( {2{\rm{a}}} \right),\left( {3{\rm{a}}} \right),\left( {4{\rm{a}}} \right)$$

$$10\,\, \le \,\,N\,\,{\mathop{\rm int}} \,\, \le \,\,99\,\,\left( 1 \right)$$
$$\left\{ \matrix{
\,N = 2M + 1,\,\,M\,\,{\mathop{\rm int}} \,\,\,\,\left( {2{\rm{a}}} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\left( { - 3 \cdot 5} \right)} \,\,\,\,\,\,\, - 15N = - 30M - 15\,\,\,\,\left( {2{\rm{b}}} \right) \hfill \cr
\,N = 3K + 2,\,\,K\,\,{\mathop{\rm int}} \,\,\,\,\left( {3{\rm{a}}} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\left( {2 \cdot 5} \right)} \,\,\,\,\,\,\,10N = 30K + 20\,\,\,\,\left( {3{\rm{b}}} \right) \hfill \cr
\,N = 5L + 4,\,\,L\,\,{\mathop{\rm int}} \,\,\,\,\left( {4{\rm{a}}} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\left( {2 \cdot 3} \right)} \,\,\,\,\,\,\,6N = 30L + 24\,\,\,\,\left( {4{\rm{b}}} \right) \hfill \cr} \right.$$
$$\mathop \Rightarrow \limits^{{\rm{sum}}\,\,\left( {{\rm{2b}}{\rm{,3b}}{\rm{,4b}}} \right)} \,\,\,\,N = 30W + 29,\,\,W\,\,{\mathop{\rm int}} \,\,\,\,\,\left\{ \matrix{
\,W = 0\,\,\, \to \,\,\,N = 29\,\,:\,\,{\rm{ok}}\,\,\left( 1 \right),\left( {2{\rm{a}}} \right),\left( {3{\rm{a}}} \right),\left( {4{\rm{a}}} \right) \hfill \cr
\,W = 1\,\,\, \to \,\,\,N = 59\,\,:\,\,{\rm{ok}}\,\,\left( 1 \right),\left( {2{\rm{a}}} \right),\left( {3{\rm{a}}} \right),\left( {4{\rm{a}}} \right) \hfill \cr
\,W = 2\,\,\, \to \,\,\,N = 89\,\,:\,\,{\rm{ok}}\,\,\left( 1 \right),\left( {2{\rm{a}}} \right),\left( {3{\rm{a}}} \right),\left( {4{\rm{a}}} \right) \hfill \cr} \right.$$
$$? = 3$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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=>

Let x be a positive integer with these properties.
Since x = 2p + 1 for some integer p, the possible values of x are x = 1, 3, 5, 7, … .
Since x = 3q + 2 some integer q, the possible values of x are x = 2, 5, 8, 11, … .
Since x = 5r + 4 for some integer 4, the possible values of x are x = 4, 9, 14, 19, … .
The first possible 2-digit number is thus 19. To find the others, note that the least common multiple of 2, 3 and 5 is lcm(2,3,5) = 30.
Thus, there are three possible 2-digit numbers with these properties:
19, 49 = 19 + 30 and 79 = 49 + 30.

Therefore, C is the answer.
Answer: C

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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, and a remainder of 4 when divided by 5?

A. 1
B. 2
C. 3
D. 4
E. 5
We see that 1, 2, and 4 are the largest remainder possible when a number is divided by 2, 3, and 5, respectively. So 1 less than the least common multiple (LCM) of 2, 3 and 5 will be an integer that has the properties mentioned in the problem. Since the LCM of 2, 3, and 5 is 2 x 3 x 5 = 30, the first 2-digit integer to have these properties is 2 x 3 x 5 - 1 = 29. Each subsequent integer is 30 more than the previous. So we have 29 + 30 = 59 and 59 + 30 = 89 as the other two 2-digit integers that have the properties. Therefore, we have a total of 3 such integers.

Answer: C

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