Company ABC has decided to distribute to its clients several shares of

Question

GMATH  practice exercise (Quant Class 17)

Company ABC has decided to distribute to its clients several shares of stocks, 1000 stocks per share. Mr. Alicante (its minor client) received 1 share, Mrs. Jung (its second minor client) received 3 shares, Mrs. Thompson (its third minor client) received 5 shares, Mr. Kumbaktoob (its fourth minor client) received 7 shares, and so on. If exactly 1,089,000 stocks were distributed by Company ABC to its clients, how many clients received those shares of stocks?

(A) 32
(B) 33
(C) 37
(D) 38
(E) 39

anshkool4u · Answer

The number of shares, to be distributed= 1089000/1000= 1089
The shares are distributed such that it forms an arithmetic progression, with first term 1, and the difference between 2 consecutive terms as 2 ( Since 1st person got 1 share,2nd got 3 shares, 3rd got 5 shares and so on).Let the shares be distributed to N people.
Now, Sum of an arithematic progression with N terms,having 1st term as A and difference between 2 terms as D is
Sum= (N/2)(2A+(N-1)D)

Here, substituting the values:
1089=(N/2)(2*1+(N-1)*2)
Gives N=33

Hence, Answer is B

fskilnik · Answer

? = N\,\,\,\,\left( {\# \,{\rm{clients}}} \right)

\left \,\,\,\,\,{{1089000} \over {1000}} = 1089\,\,\, = \,\,1 + 3 + 5 + 7 + \ldots + \left( {2N - 1} \right)\,\,\,\,\,\,\left( * \right)

\left( * \right)\,\,::\,\,\,\,1 = 2 \cdot \boxed1 - 1\,\,,\,\,\,3 = 2 \cdot \boxed2 - 1\,\,,\,\,\,5 = 2 \cdot \boxed3 - 1\,\,,\,\,\, \ldots

1 + 3 + 5 + 7 + \ldots + \left( {2N - 1} \right)\,\,\,\mathop = \limits^{{\rm{Arith}}{\rm{.}}\,{\rm{Seq}}.} \,\,\,\,\left \, \cdot N\,\,\, = \,\,\,{N^2}

{N^2} = 1089 = {3^2} \cdot {11^2}\,\,\,\,\,\mathop \Rightarrow \limits^{N\, > \,0} \,\,\,\,? = N = 33

The correct answer is (B).

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

Zoom96 · Answer

n^2 = 1089 = 3^2 * 11^2 <=> n= 3*11 = 33

ArtyomYom · Answer

Company ABC has decided to distribute to its clients several shares of stocks, 1000 stocks per share. Mr. Alicante (its minor client) received 1 share, Mrs. Jung (its second minor client) received 3 shares, Mrs. Thompson (its third minor client) received 5 shares, Mr. Kumbaktoob (its fourth minor client) received 7 shares, and so on. If exactly 1,089,000 stocks were distributed by Company ABC to its clients, how many clients received those shares of stocks?

(A) 32
(B) 33
(C) 37
(D) 38
(E) 39

1089000 / 1000 = 1089

Next check divisibility of our options on 1089

1089 clearly doesn't divisible by A and D, coz they are even

Try B 33 = 3*11, so 1+0+8+9 = 18, 18/3 = 6 - ok
0+9-(1+8) = 0 - ok B is answer
**If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11.**

Option C - 37, here is no shortcuts, so need to divide 37 on 1089 ; 1089 / 37 = 29,.... => wrong

Option E - 39 = 3 * 13, need to divide 13 on 1089 ; 1089 / 13 = 83,.... => wrong

bumpbot · Answer

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Company ABC has decided to distribute to its clients several shares of

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