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Please feel free to change the subject lines. They are not my problems. However, I did not see any copyright restrictions where I found them so I personally don't mind if you use them.
Wow thank you.
Oops is it possible to change the forum topic to Integer Partitions?
Yes am actually learning python currently. Thank you for your time bobbym.
Yes that is exactly how the question is set. You have been very helpful, thank you.
bobbym,
Thank you.
I got 9604
X(number of term)=99-2+1=98
Y(number of term)=99-2+1=98
XY(number of unique integer) =98×98=9604
Please advice me how to go about this question.
Also I noticed you responded to the other forum questions titled Combinations and Permutations. Thank you.
Can you teach me how to reason thru those questions?
Thank you so much guys.
Could someone please teach me how to do this the hard way? i.e. pen & paper?
A book recommendation or any other online source will also be appreciated.
If you have four coins, you can split them into piles in five different ways:
s(4)=5. If s(x) is the number of ways to split x coins, find the minimum value of x for which s(x) is divisible by 100 000 (one hundred thousand).
A• 12 547
B• 11 378
C• 11 224
A property developer can build four different house sizes on a street. The first house size takes up one lot, the second house size takes up two lots, the third house size takes up three lots and the fourth house size takes up four lots. If a street measures four lots, the developer can develop the street in eight different combinations:
In how many different ways can the developer build the houses if the street measures 20 lots?
A• 387 954
B• 253 971
C• 283 953
In South Africa the currency is made up of Rands, R, and cents, c, and there are six coins currently in circulation:
1c, 2c, 5c, 10c, 20c, 50c, R1, R2, R5.
It is possible to make R10 in the following way:
1 × R5 + 1 × R2 + 2 × R1 + 1 × 50c + 1 × 20c + 2 × 10c + 1 × 5c + 1 × 2c + 3 × 1c
How many different ways can R10 be made using any number of coins?
A• 335 868 921
B• 41 789 014
C• 265 532 101
For all integer combinations xy for 2 ≤ x ≤ 6 and 2 ≤ y ≤ 4:
22=4, 23=8, 24=16
32=9, 33=27, 34=81
42=16, 43=64, 44=256
52=25, 53=125, 54=625
62=36, 63=216, 64=1296
The sequence generated contains 15 unique terms:
4, 8, 9, 16, 25, 27, 36, 64, 81, 125, 216, 256, 625, 1296
How many unique integers are in the sequence generated by xy for 2 ≤ x ≤ 99 and 2 ≤ y ≤ 99?
A• 9 045
B• 7 829
C• 7 090
Thank you so much.
bob bundy,
The question arose from a graduate management test.
These are the answer choices given:
A. 1 880
B. 1 043
C. 1 309
I do not have the official answer.
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