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Bunuel
When the positive integer n is divided by 7, the quotient is q and the remainder is 4. When 2n is divided by 7, the remainder is 1 andthe quotient, in terms of q, is


A. q/2

B. q/2 + 1

C. 2q

D. 2q+1

E. 2q+2

Let \(n = 11\), So \(q = 1\) and \(r = 4\)

When \(2n\) is divided by \(7\) then we have \(\frac{22}{7} = 7*3(q) + 1(r)\)

Thus, the value of quotient in terms of \(q(3)\) = \(2*1(q) + 1\), Answer must be (D)
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Bunuel
When the positive integer n is divided by 7, the quotient is q and the remainder is 4. When 2n is divided by 7, the remainder is 1 and the quotient, in terms of q, is


A. q/2

B. q/2 + 1

C. 2q

D. 2q+1

E. 2q+2

IMO D

Plug in some values n = 11, q = 1, remainder = 4

Now n becomes 2n = 22 here q = 3, remainder = 1

So only D satisfies the above relation

2*1 + 1 = 3
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Bunuel
When the positive integer n is divided by 7, the quotient is q and the remainder is 4. When 2n is divided by 7, the remainder is 1 and the quotient, in terms of q, is


A. q/2

B. q/2 + 1

C. 2q

D. 2q+1

E. 2q+2


Two ways..

(I) Algebraic
the positive integer n is divided by 7, the quotient is q and the remainder is 4 MEANS n=7q+4.
2n is divided by 7, the remainder is 1 and the quotient, in terms of q..2n=7x+1.
substitute \(n=7q+4, .....2(7q+4)=7x+1......14q+8=7x+1......7x=14q+8-1=14q+7.......x=2q+1\)
D

(II) Substitute n as some value..
take q as 2, so n=7*2+4=18..
so 2n = 2*18=36
36=7x+1.......7x=35..x=5

Substitute q as 2 and see which choice gives you 5 ..
2q+1=2*2+1=5

so D
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Let quotient interms of q =?
Now since dividend is n ,divisor is 7 ,quotient is q and remainder being 4 can be in this form ,n=7q+4.......(1)
Now second statement says ,
2n=?+1.....(2)
Multiply eqn.....(1) by 2
2n=14q+8
2n=?+1 —> 14q+8=?+1 —> 14q+7=? —> 7(2q+1)=?
Haha so there we have it divisor*quotient ,so quotient interms of q is (2q+1)
Answer is kindly D
Hope this helps .

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Hello everyone!

Could someone please tell me if my logic is good or it was just a coincidence?

"the remainder is 1.."

So in my head, I just canceled out A, C and E because they do not hold a remainder of 1.
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Bunuel
When the positive integer n is divided by 7, the quotient is q and the remainder is 4. When 2n is divided by 7, the remainder is 1 and the quotient, in terms of q, is


A. q/2

B. q/2 + 1

C. 2q

D. 2q+1

E. 2q+2

We can create the equations:

n/7 = q + 4/7

n = 7q + 4

and

2n/7 = z + 1/7

2n = 7z + 1

Multiplying the first equation by 2, we have:

2n = 14q + 8

Setting the two equations equal to each other, we have:

14q + 8 = 7z + 1

14q + 7= 7z

2q + 1 = z

Alternate Solution:

Since the remainder is 4 when n is divided by 7, we see that n could be 11, 18, 25, 32, …

For simplicity’s sake, let’s let n = 11. When 11 is divided by 7, the quotient q is 1, with a remainder of 4.

Now, we see that 2n = 22. When 22 is divided by 7, the quotient is 3, with a remainder of 1.

The original quotient q is 1, and the new quotient is 3. Of the answer choices, only D. 2q + 1 gives us an answer of 3 when q = 1.

Answer: D
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