If x is a positive integer, r is the remainder when x is divided by 4,

x = 4a+r

i.e. \(r_{Max} = 3\)

x = 9a+R

i.e. remainders for divisor 49may be {0, 1, 2, 3, 4, 5, 6, 7, 8}[/b]

i.e. \(R_{Max} = 8\)

\(r^2 + R = 3^2 + 8 = 17\)

Answer: Option C

\(x\) — positive integer
\(x= 4m+ r \)
\(x = 9n + R \)

r could be maximum 3
R could be maximum 8
—>\( r^{2} + R = 9+ 8 = 17 \)

Answer (C)

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Lets assume x to be a few different numbers - 1,2,3,4,5,6,7,8,9 (Considering many numbers just to clarify the concept)

When x is divided by 4

1/4 - The remainder will be 1
2/4 - The remainder will be 2
3/4 - The remainder will be 3
4/4 - The remainder will be 0 (4 is directly divisible by 4)
5/4 - The remainder will be 1
6/4 - The remainder will be 2
7/4 - The remainder will be 3
8/4 - The remainder will be 0 (8 is directly divisible by 4)
9/4 - The remainder will be 1

If we notice, the cycle keeps on repeating and we learn that the highest remainder is 3 when x si divided by 4

Similarly when X is divided by 9
1/9 - The remainder will be 1
2/9 - The remainder will be 2
.....
The highest remainder can be
8/9 - Where the remainder can be 8 After which the remainder cycle will again start from 0

Now, to answer the question asked
r has a maximum remainder of 3 and R has a maximum remainder of 8

So r^2 + R = 3^2 + 8 = 17

Sadly... I got the same doubt and quickly took lcm of no. S and worked on no. S around it take 35 that no. Will satisfy both.

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Tanchat

"Can we always use this logic?" is a very tricky question.
Whether I can use particular logic depends on what is given to me. I can change the question a little bit and the logic I will use changes.

Here we have to maximise r^2 + R. Maximum value of r is 3 and that of R is 8 i.e. Divisor - 1. So remainder in both cases should be -1 (if you are comfortable with negative remainders).
Then the number LCM - 1 will give remainder -1 in both cases i.e. remainder 3 when divided by 4 and remainder 8 when divided by 9.
Since LCM of 4 and 9 is 36, when 36 - 1 = 35 is divided by 4, remainder is 3 and when it is divided by 9, remainder is 8.

Note that you have written r^2 - R above (which I assume is a typo).
But if this were actually given, to maximise this, we would need to maximise r (= 3) and minimise R (= 0).

Then we would need x such that division by 4 gives remainder 3 and division by 9 gives remainder 0.

x = 4a + 3
x = 9b
Here we don't have a common remainder so we will need to find such value of x with hit and trial. We find that the first such value of x would be 27.
Then maximum value of r^2 - R is 3^2 - 0 = 9.

Now think how the question changes when you obtain r when divided by 8 and R when divided by 12. What will be the maximum value of r^2 - R? Will it be 7^2 - 0? Is there a value of x that will give 7 upon division by 8 but will be divisible by 12? No because (8a + 7) is an odd number. So then can R be 0 or 1 or 2 or 3 or 4? Think about it.

Hello,

It does make sense. But in peer pressure of less time, one might not able to get quickly to series of those numbers that will satisfy both conditions. So, better pay attention to 'r' and 'R' and solve it quickly.

Thanks

Hi

There is no restrictions on values of x here, that is you can take any value of x.

So, yes there will always be a number fitting in for x. Here, when you subtract 1 from LCM(4,9) and its multiples will fit in.
For example: 36-1 or 35.=> r is 3 and R is 8. Similarly 72-1 or 71

Given:
1. x is a positive integer
2. r is the remainder when x is divided by 4
3. R is the remainder when x is divided by 9

Asked: What is the greatest possible value of r^2 + R ?

2. r is the remainder when x is divided by 4
Maximum value of r = 3
3. R is the remainder when x is divided by 9
Maximum value of R = 8
Maximum value of r^2 + R = 3^2 + 8 = 9 + 8 = 17

IMO C

ETA: Per below - I was wrong - thanks all!

Don't these answers mistakenly infer that x is actually two different values? I read the question that x must be the same for both conditions, so the max value is actually 15 (for example, when x = 15, r = 3 and R = 6).

There is no single number x where r=3 and R=8.

I think the question is misleading.

Hi experts,

This question is easy if we just focus on maximizing the two remainders 'r' and 'R'. However, the question implies that 'x' hast to be the same positive integer for both the divisors 4 and 9. How/why did you experts choose to completely overlook this and just focus on maximizing the remainders?

Agree.

Bunuel - for this question, do we have to test 3 and 8, or can we just assume 3 and 8 since they are the highest?

i.e. in a similar question, might we have a case where the highest possible remainders for each divisor do NOT both work for a given number and we must narrow down a pair of remainders that do work for a given number?

x = 4k + r
x = 9p + R

Say x = 7 divided by 4 then we get remainder as 3. [Other values of x = 5, 6 will give remainder as 1 and 2 respectively]
Say x = 17 divided by 9 then we have remainder as 8.[Other values of x from 10 to 16 will give remainder less than 8]

=> r = 3 and R = 8. Therefore, \(3^2 + 8 = 17\)

Answer C

Hi MathRevolution


Finding a dividend that gives us a remainder of 3 when divided by 4 and a remainder of 8 when divided by 9 is not difficult.

The question says, "....r is the remainder when x is divided by 4, and R is the remainder when x is divided by 9..". Therefore, doesn't the dividend 'x' have to be the same for both the divisor 4 and 9?

Hi experts,

Bunuel chetan2u
KarishmaB

For this type of question, can we always use this logic : the greatest remainder is always divisor - 1?
In this case, can we safely assume that 4-1 and 9-1 can make the greatest of r^2 - R

or we have to always check that there is at least a possible value of x that can serve : x = 4p + 3 and x = 9q + 8.

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