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Number System questions for IPMAT, with solutions

10 free Number System practice questions for IPMAT Indore, Rohtak and JIPMAT, 4 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Number System on Learn.

1. Question 1Type the answerWritten by IPMATwayNumber Bases

   The decimal equivalent of the octal number $(273)_8$ is

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   Answer: 187

   In base $8$ the place values from the right are $1$, $8$ and $8^2=64$. So $(273)_8=2\times64+7\times8+3\times1=128+56+3=187$.

2. Question 2Type the answerWritten by IPMATwayTypes of Numbers

   The product of two consecutive even numbers is $624$. Their sum is

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   Answer: 50

   Let the numbers be $x$ and $x+2$. Then $x(x+2)=624$, i.e. $x^2+2x+1=625$, so $(x+1)^2=625$ and $x+1=25$. The numbers are $24$ and $26$ (check $24\times26=624$), and their sum is $50$.

3. Question 3Type the answerWritten by IPMATwayFactorials and Trailing Zeros

   If $146!$ is divisible by $6^{n}$, then the maximum value of $n$ is

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   Answer: 70

   $6^n=2^n3^n$. Legendre: $v_3(146!)=\lfloor146/3\rfloor+\lfloor146/9\rfloor+\lfloor146/27\rfloor+\lfloor146/81\rfloor=48+16+5+1=70$, far fewer than the $143$ factors of $2$. Hence $n_{\max}=70$.

4. Question 4Multiple choiceAsked in JIPMAT 2023Factors and Divisors

   Amicable numbers are a pair of distinct natural numbers $(a, b)$ such that the sum of the proper divisors of $a$ equals $b$ and the sum of the proper divisors of $b$ equals $a$. Given that $(220, y)$ is a pair of amicable numbers, $y=$ ________.

   1. A.$225$
   2. B.$284$
   3. C.$504$
   4. D.$768$
   Show the answer and solution

   Answer: B. $284$

   Sum of all divisors of $220$: $(1+2+4)(1+5)(1+11)=504$. Subtract $220$: $284$. Check: divisors of $284=2^2\cdot71$ sum to $7\times72=504$, and $504-284=220$.

5. Question 5Type the answerWritten by IPMATwayInteger Solutions of Equations

   A test has $80$ questions. Each correct answer earns $1$ mark, each wrong answer costs $\dfrac12$ mark and an unattempted question earns $0$. A student left $10$ questions unattempted and scored $40$ marks. The number of questions he answered correctly is

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   Answer: 50

   $10$ questions were left, so $80-10=70$ were attempted: $c+w=70$. The score is $c-\frac{w}{2}=40$, i.e. $2c-w=80$. Adding the two equations: $3c=150$, so $c=50$ correct (and $w=20$ wrong; check $50-10=40$).

6. Question 6Type the answerWritten by IPMATwayRemainders

   The remainder when $1!+2!+3!+\dots+100!$ is divided by $40$ is

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   Answer: 33

   $40=8\times5$. $5!=120=3\times40$, and every factorial from $5!$ on is a multiple of $5!$, hence of $40$. So only $1!+2!+3!+4!=33$ matters. $33<40$, so the remainder is $33$.

7. Question 7Multiple choiceAsked in IPMAT Rohtak 2020Simplification and Approximation

   What is cube root of $55555$?

   1. A.$38.17$
   2. B.$38.16$
   3. C.$38.15$
   4. D.$38.14$
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   Answer: B. $38.16$

   $38^3=54872$ and $39^3=59319$, so the root is between $38$ and $39$. Tangent estimate: $38+\frac{55555-54872}{3\cdot38^2}=38+\frac{683}{4332}\approx38.158$. Check: $38.15^3\approx55524$ and $38.16^3\approx55568$, so the root is $38.157\ldots$, which rounds to $38.16$.

8. Question 8Multiple choiceAsked in IPMAT Indore 2021Unit and Last Digits

   The unit digit in $(743)^{85}-(525)^{37}+(987)^{96}$ is ________

   1. A.$9$
   2. B.$3$
   3. C.$1$
   4. D.$5$
   Show the answer and solution

   Answer: A. $9$

   Only unit digits matter. $3$ has cycle $3,9,7,1$; $85=4\times21+1$ gives $3$. Powers of $5$ end in $5$. $7$ has cycle $7,9,3,1$; $96$ is a multiple of $4$ so $1$. $3-5+1=-1$, i.e. the number ends in $9$.

9. Question 9Multiple choiceAsked in IPMAT Indore 2022Divisibility Rules

   If the five-digit number $abcde$ is divisible by $6$, then which of the following numbers is not necessarily divisible by $6$?

   1. A.$edcba$
   2. B.$eee$
   3. C.$bbadcacede$
   4. D.$cdbae$
   Show the answer and solution

   Answer: A. $edcba$

   From $abcde$ divisible by $6$: $e$ is even and $a+b+c+d+e\equiv0\pmod3$. $eee$, $bbadcacede$ and $cdbae$ all end in $e$ and have digit sums $3e$, $2(a+b+c+d+e)$, $a+b+c+d+e$ - multiples of $3$. $edcba$ keeps the digit sum but ends in $a$, so it fails whenever $a$ is odd.

10. Question 10Type the answerWritten by IPMATwayDigits and Number Forms

    How many four-digit positive integers have the product of their digits equal to $24$?

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    Answer: 64

    No digit can be 0. List multisets of four digits from 1 to 9 with product 24: $\{1,1,3,8\}$, $\{1,1,4,6\}$, $\{1,2,2,6\}$, $\{1,2,3,4\}$, $\{2,2,2,3\}$ ($\{1,1,2,12\}$ is invalid as 12 is not a digit). Arrangements: $\frac{4!}{2!}=12$ for each of the first three; $4!=24$ for $\{1,2,3,4\}$; $\frac{4!}{3!}=4$ for $\{2,2,2,3\}$. Total $12+12+12+24+4=64$.