## 46th SSC CGL level Solution Set, 7th on topic Number System

This is the 46th solution set of 10 practice problem exercise for SSC CGL exam and 7th on topic Number System. Students must complete the corresponding question set in prescribed time first and then only refer to the solution set for extracting maximum benefits from this resource.

In MCQ test, you need to deduce the answer in shortest possible time and select the right choice.

Based on our analysis and experience we have seen that, for accurate and quick answering, the student

- must have complete understanding of the basic concepts in the topic area
- is adequately fast in mental math calculation
- should try to solve each problem using the basic and rich concepts in the specific topic area and
- does most of the deductive reasoning and calculation in his or her head rather than on paper.

Actual problem solving is done in the fourth layer. You need to use **your problem solving abilities** to gain an edge in competition.

If you have not taken this test yet, you may refer to the * 46th SSC CGL level Question Set on Number System 7* to take the test before going ahead with its solution here.

### 46th solution set- 10 problems for SSC CGL exam: 7th on topic Number System - time 12 mins

**Problem 1.**

Two numbers 7655 and 11284 when divided by a third number of three digits, the remainder left in both cases is same. The sum of the digits of the three digit number is then,

- 9
- 11
- 10
- 8

** Solution 1 : Problem analysis and execution:**

In the two divisions, when the divisors as well as the remainders are same, we conclude - it will easily be possible to reduce the problem to a problem of factorization by using Euclid's fundamental general relation of dividend, divisor, quotient and remainder in a division,

$a=bq+r$, where $a$ is dividend, $b$ is the divisor, $q$ is the quotient and $r$ the remainder.

In two divisions according to Euclid's relation of division, we have,

$7655=bq_1+r$, where $b$ is the same divisor, $r$ is the same remander and $q_1$ is the quotient in this first division.

Similarly in the second division we have,

$11284=bq_2+r$, here $q_2$ is assumed to be the quotient.

By subtracting the first equation from the second we now eliminate the same remainder,

$3629=b(q_2-q_1)$.

Both $b$ and $(q_2-q_1)$ are then the two factors of 3629.

When we examine 3629, we find that $b$ being a number of three digits, even if we divide 3629 by 100, the smallest three digit number, the result is only just more than 36.

The problem is then transformed to finding a prime factor of 3629 less than 36. Note that instead of finding $b$, the target number directly, we have adopted the * indirect approach* of finding the second factor $(q_2-q_1)$. This factor being smaller and within an estimated value, it will be a shorter path to finding this smaller factor which in turn will give us the value of $b$ automatically as the quotient of the division.

When applying the standard method of factorization, test with 2, 3 (sum of digits divisible by 3) and 5 fail immediately. Mentally we divide by 7 which also fails. Testing with 11 as a factor using 11 divisibility rule (difference of sum of alternate digits either 0 or divisible by 11) failing we test divisibility of 3629 with 13, and then 17 by direct mental division both of which fail.

But with the next prime 19, we get success,

$3629=191\times{19}$.

Our target three digit number we have found indirectly as 191 and the sum of its digits is 11.

**Answer:** b: 11.

**Key concepts used:** Euclid's fundamental relation of a general division -- elimination of the remainder to transform the problem to a problem of factorization -- problem transformation -- assessment of the complexity of factorization -- standard factorization method using efficient divisibility rules.

We have used all the basic concepts and techniques that were the most suitable for solving the problem. Deductive reasoning was the driving mechanism in the whole process of solving the problem quickly.

**Problem 2.**

The value of $(0.34\overline{68}+0.17\overline{32})$ is,

- $0.52\overline{01}$
- $0.\overline{51}$
- $0.52$
- $0.5\overline{1}$

**Solution 2 : Problem analysis and execution**

To show the mechanism of solution, we will write two more repeating sequences of two digits for both the numbers involved in the proble,

$0.3 4 6 8 6 8 6 8\overline{68}......$

$0.1 7 3 2 3 2 3 2\overline{32}......$

--------------------------------------

$0.5 2 0 1 0 1 0 0.......$

The third pair of 68 and 32 from the left add up to 100, with a carry over of 1. This carry over flows through the two more pairs of repeating digits on the left into the non-repeating digit area, and also results in $01$ as a seemingly repeating pattern.

In other words, first pair of repeating digits from the right add up to 100, while all the repeating digit pairs on its left add up to 101, consisting of carry over of 1 and a two digit sequence of 01.

This is clear, but what the final result will be is not so clear.

#### Use of infinity in two ways

In a non-terminating repeating digit sequence the repeating pattern of say, $68$ continues to the right endlessly to infinity. The same is true for the second repeating digit pattern of 32.

In the first instance we will assume infinity to end at a point where the absolute rightmost two pairs, $68$ and $32$ add up to 100 resulting in a two digit result of $00$ and a carry of 1. This is the only time a $00$ is produced and being the rightmost zeros these can be ignored. The carry instead will generate a sequence of $01$ all through to the left of infinity and finally will spill over into the non-repeating digit area.

So we can say that the result of addition is, $0.52\overline{01}$. In this process of getting the final result we are using infinity as a no-end point.

**Answer:** Option a : $0.52\overline{01}$ .

**Key concepts used:** Rich concept of addition of non-terminating repeating digits -- infinity concept -- use of infinity -- basic addition concept.

**Problem 3.**

If $p$ and $q$ are positive integers and $(p-q)$ is an even number, then $(p^2-q^2)$ will always be divisible by,

- 4
- 6
- 12
- 8

**Solution 3 : Problem analysis and execution:**

From the concept of odd and even numbers we know, if $p-q$ is an even number,

- either both $p$ and $q$ are odd numbers, or,
- both $p$ and $q$ are even numbers.

There is no third possibility.

So it follows immediately that $p+q$ also is an even number having a factor of 2.

Finally then, $p^2-q^2=(p-q)(p+q)$ will have two numbers of 2, that is, 4 as a factor.

**Answer:** Option a: 4.

**Key concepts used:** *Odd and even number concept related to addition and subtraction -- looking at the sum and difference of the two numbers as two factors of $p^2-q^2$.*

**Problem 4.**

3957 is added to 5349 and 7062 is subtracted from the resulting sum. The final result is then not divisible by,

- 3
- 4
- 7
- 11

**Solution 4 : Problem analysis and execution:**

We decided that the quickest and most accurate method will be to add the two numbers first and then subtract the third number from the sum following the manual methods of addition and subtraction taught in school. The result came out quickly to be 2244. Now it is a smple testing of the choice values as factors.

Integer sum of 2244 is divisible by 3. Last two digits being divisible by 4, the number is divisible by 4. It is apparent that the number is divisible by 11. Finally division by 7 fails.

**Answer:** c: 7.

**Key concepts used:** Quick addition and subtraction -- divisibility rules.

**Problem 5.**

If the sum of digits of any integer lying between 100 and 1000 is subtracted from the number, the result will always be divisible by,

- 9
- 6
- 2
- 5

**Solution 5 : Problem analysis and execution:**

First we randomly assume a 3 digit number, 469 and subtracting its sum of digits from itself, obtain 450 which is divisible by 2, 5, 6, and 9 all. Next we again assume a random 3 digit number, 878 getting derived result 855, which is divisible by 5 and 9. So with this test we have been able to eliminate the other two choice values 2 and 6.

Next we assume randomly 423, getting derived result, 414, divisible by 9, 2 and 6, but not 5.

So by applying exclusion principle on the three results we conclude, the result will always be divisible by only 9.

**Answer:** Option a: 9.

**Key concepts used:** Enumeration technique -- logic analysis -- **inclusion exclusion analysis.**

#### Formal method

Any three digit number is assumed to be, $abc$. Expanding it by place value structure,

$abc=100a+10b+c$.

We are to subtract the sum of the digits, that is, $(a+b+c)$ from the number,

$abc-(a+b+c)$

$=100a+10b+c-(a+b+c)$

$=99a+9b$, always divisible by 9.

If you ask us, we will always choose this * conceptual approach* rather than the enumeration technique of hit and try.

Furthermore, this is an example of * many ways technique* of solving a problem in more than one method.

**Repeated practice of this technique enhances your problem solving skill by increasing your ability to new approaches.****Key concepts** **used:** * Conceptual approach* --

*-- breaking up of a number --*

**Place value structure**

**many ways technique.**#### Problem 6.

How many numbers between 400 and 900 are divisible by 4, 5 and 6?

- 8
- 7
- 9
- 10

**Solution 6 : Problem analysis and execution:**

A number to be divisible by 4, 5 and 6, it must be divisible by the LCM of the numbers which is 60 in this case. We derive it mentally.

The problem is transformed to, finding how many numbers between 400 and 900 are divisible by 60. Note carefully at this point only, in the range, both 400 and 900 are not included.

As a first step, we will land on to 420 as the first multiple of 60 just crossing over into the range. For safety and the range not being large we will use 180 which is 3 times 60 as a **jump step.**

Adding the jump step 180 to 420 we get 600 and count four 60s within the range. Addition of another 180 gives result 780 and count $4+3=7$.

We can see that 900 is just 120 that is two 60s away. But as the range boundary is not to be included, we can add only one more 60 to make the count as 8.

**Answer:** Option a : 8.

**Key concepts used:** Clear problem definition -- problem transformation -- LCM -- ** Jump step technique** -- range estimation.

** Problem 7.**

When a number is divided by 56, the remainder left is 29. What will be the remainder when the number is divided by 7?

- 6
- 3
- 1
- 2

**Solution 7 : Problem analysis and execution:**

Using Euclid's division formula in the first instance we get,

$n =56q+29$, where $n$ is the number being divided by 56, $q$ the quotient and remainder 29.

We need to divide both sides of the equation by 7 to get the desired remainder on dividing $n$ by 7. The remainder in this case will be generated only from the second term of the division, that is, $29\div{7}$ which is, 1. This is so as 7 is a factor of 56 and will generate a remainder 0 from the first term of the RHS.

**Answer:** Option c: 1.

** Key concepts used:** Euclid's division relation -- remainder concepts.

** Problem 8.**

64329 is divided by a certain number getting 175, 114 and 213 as three successive remainders. The divisor is then,

- 296
- 234
- 184
- 224

**Solution 8 : Problem analysis and execution:**

At first we were unable to understand how three remainders can be generated out of a single division. After a moment though the meaning of the phrase * successive remainders* dawned on us - the problem asks us to use the concept of manual division that we learned in school. There was thus a bit of a hesitation in

*itself.*

**problem definition**In the first step of division as 643 is divided by the three digit divisor leaving a remainder of 175, the unknown divisor must divide $(643-175)=468$ fully. In other words the unknown divisor must be a factor of 468. It follows from the basic concept of division process.

In the second step, the new number 1752 is divided by the divisor leaving a remainder of 114. By the same logic then we can conclude, the unknown divisor is a factor of $(1752-114)=1638$.

In the third step of division, 1149 is divided leaving a remainder of 213. So the unknown divisor is also a factor of, $(1149-213)=936$.

Collecting the results from the three successive steps of division we get,

The unknown divisor is a three digit common factor of 468, 1638 and 936.

Just a quick look at the choice values identify 234 as this unknown common factor. $7\times{234}=1638$ we could derive mentally, other two are trivial.

**Answer:** Option b: 234.

**Key concepts used:** Manual division concept -- remainder concept -- common factor concept -- *common factor identification by pattern.*

**Problem 9.**

A number when successively divided by 4 and 5 leaves remainders 1 and 4 respectively. The remainders left when the number is divided by 5 and 4 successively will be,

- 1, 4
- 2, 1
- 2, 3
- 3, 2

**Solution 9 : Problem analysis and execution:**

If $n$ be the number and quotient of first division be $q_1$,

$n=4q_1+1$.

Similarly for the second division the quotient of the first division is divided by 5 so that we get,

$q_1=5q_2+4$,

Or, $4q_1=20q_2+16$.

Substituting in the first equation,

$n=20q_2+17$

If we divide $n$ by first 5 and second 4, finally it will still be equivalent to dividing $n$ by 20 getting a remainder of 17. The only difference will be in the break up of the remainders in each step.

In this second case, dividing $n$ by 5 results in a remainder generated by dividing 17 by 5 which is 2. The quotient of this first division by 5 being, $4q_2+3$ taking away 15 from 17, and the second division of $4q_2+3$ by 4 will generate a remainder of 3.

**Answer:** Option c: 2, 3.

**Key concepts used:** Basic concepts of a division, forming quotient and remainder -- * successive division concept* -- Euclid's division formula.

** Problem 10.**

If two numbers are each divided by the same number the remainders are respectively 4 and 3. If again the sum of the numbers are divided by the same divisor, the remainder now is 2. The divisor is,

- 3
- 9
- 5
- 7

**Solution 10 : Problem analysis and execution:**

By Euclid's division relation the first division gets us,

$m=bp+4$, where $m$ is assumed as the first number, $p$ the first quotient and $b$ the same divisor that we have to find out.

Similarly in the second division we get,

$n=bq+3$, where $n$ is the second number and $q$ the second quotient.

Summing the two we get,

$(m+n)=b(p+q)+7$.

This is a representation of division of $(m+n)$ by the same divisor. As the first term on the RHS will not generate any remainder, $b$ itself being a factor of the term, the second term 7 only will generate a remainder by division by the same divisor.

The final remainder of the third division being 2, the divisor must be, $7-2=5$.

**Answer:** Option c: 5.

**Key concepts used:** Euclid's division relation -- remainder concepts -- factor concepts.

**Note:** In almost all the problems a clear understanding of the very basic concepts was called for.

#### Valuable guidelines with links to resources on SSC CGL

**7 steps for sure success in SSC CGL tier 1 and tier 2 Competitive tests**

#### Question set corresponding to this solution set

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#### Tutorials

**Numbers and Number system and basic mathematical operations**

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**Basic and Rich Algebra concepts for elegant solutions of SSC CGL problems - **** though on Algebra, it should be useful.**

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