Find the maximum value of n such that 61 x 42 x 26 x 81 x 64 x 49 x 47 x 51 x 63 x 16 x 67 is perfectly divisible by 56^n Option 1: 6 Option 2: 5 Option 3: 4 Option 4: 3

Answer: 4. Solution: Break 56 into primes: 56=2^3x7 So we only need to know how many 2s and how many 7s the big product has., Number, 2s it gives, 7s it gives, 42=2x3x7, 1, 1, 26=2x13, 1, 0, 64=2^6, 6, 0, 16=2^4, 4, 0, 49=7x7, 0, 2, 63=7x9, 0, 1, The remaining numbers 61, 81, 47, 51 and 67 contain neither 2 nor 7. Total 2s=1+1+6+4=12 Total 7s=1+2+1=4 Every single 56 needs three 2s and one 7. The 2s allow 12/3=4 such groups. The 7s allow 4/1=4 such groups. We must take the smaller of the two, so n=4

Number System

Divisibility Rules

Hard

Find the maximum value of $n$ such that

$61 \times 42 \times 26 \times 81 \times 64 \times 49 \times 47 \times 51 \times 63 \times 16 \times 67$ is perfectly divisible by $56^n$

✅ Correct Option: 3

Break 56 into primes:

$56=2^3\times7$

So we only need to know how many 2s and how many 7s the big product has.

Number	2s it gives	7s it gives
$42=2\times3\times7$	1	1
$26=2\times13$	1	0
$64=2^6$	6	0
$16=2^4$	4	0
$49=7\times7$	0	2
$63=7\times9$	0	1

The remaining numbers $61$ , $81$ , $47$ , $51$ and $67$ contain neither 2 nor 7.

$\text{Total 2s}=1+1+6+4=12$

$\text{Total 7s}=1+2+1=4$

Every single 56 needs three 2s and one 7.

The 2s allow $\dfrac{12}{3}=4$ such groups.

The 7s allow $\dfrac{4}{1}=4$ such groups.

We must take the smaller of the two, so

$n=4$

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Find the maximum value of nnn such that 61×42×26×81×64×49×47×51×63×16×6761 \times 42 \times 26 \times 81 \times 64 \times 49 \times 47 \times 51 \times 63 \times 16 \times 6761×42×26×81×64×49×47×51×63×16×67 is perfectly divisible by 56n56^n56n

Solution

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JIPMAT Past Year Questions (Topic-Wise)

Find the maximum value of $n$ such that

$61 \times 42 \times 26 \times 81 \times 64 \times 49 \times 47 \times 51 \times 63 \times 16 \times 67$ is perfectly divisible by $56^n$