Gauss Contest
Which of the following is a multiple of 7?
(A) 75 (B) 76 (C) 77 (D) 78 (E) 79
Solution 1 A number is a multiple of 7 if it is the result of multiplying 7 by an integer. Of the answers given, only 77 results from multiplying 7 by an integer, since 77 = 7 × 11.
Solution 2 A number is a multiple of 7 if the result after dividing it by 7 is an integer. Of the answers given, only 77 results in an integer after dividing by 7, since 77 ÷ 7 = 11.
Answer: (C) 77
Mete sleeps daily for 3 times as many hours as he is awake. For how many hours does Al sleep daily?
A) 6 B) 9 C) 12 D) 18
Mete sleeps daily for 3/4 of the day. Therefore, Al sleeps for 18 hours.
D) 18
What is a multiple of 7
A)90 B)91 C)92 D)93 E)94
90/7 = 12R6 91/7 = 13 92/7 = 13R1 93/7 = 13R2 94/7 - 13R3
Answer is B)91
What is the first number containing an "i"
5 (five)
Jesse pays $2.25 to take the bus. Ariel pays $3.00 to take the bus. If they each take the bus 20 times, how much less would Jesse pay than Ariel in total?
(A) $25 (B) $10 (C) $15 (D) $45 (E) $60
Solution 1 Since Jesse pays $2.25 to take the bus, then 20 trips on the bus would cost Jesse 20×$2.25 = $45. Since Ariel pays $3.00 to take the bus, then 20 trips on the bus would cost Ariel 20×$3.00 = $60. If they each take the bus 20 times, then in total Jesse would pay $60 − $45 = $15 less than Ariel.
Solution 2 Since Jesse pays $2.25 to take the bus, and Ariel pays $3.00 to take the bus, then Jesse pays $3.00 − $2.25 = $0.75 less than Ariel each time they take the bus. If they each take the bus 20 times, then in total Jesse would pay 20 × $0.75 = $15 less than Ariel.
Answer: (C) $15
The product of four 4s equals the sum of _?_ 4s.
A) 4 B) 3*4 C) 43 D) 44
The product of four 4s = 256 = 4*64; this is the sum of 64 4s.
C) 43
If Tom(me) tosses a coin, rolls a dice and spins a equal four sided wheel, what is the chance that he gets heads, a 6, and anything on the wheel?
A)1/48 B)1/24 C)1/12 D)1/6 E) 50%
Since the spinner doesn't matter due to it can land on anything, it is 1/2 chance for heads times 1/6 chance for a 6, therefore being 1/12
Answer is C)1/12
What is the first number containing a "d"
100 (hundred)
Mr. A chooses two different items for a snack. His choices are a snickers bar, some smarties, a banana, and seaweed. How many different pairs of snacks could he choose?
(A) 3 (B) 4 (C) 5 (D) 6 (E) 7
Mr. A could choose from the following pairs of snacks: snickers bar and smarties, snickers bar and banana, snickers bar and seaweed, smarties and banana, smarties and seaweed, or banana and seaweed. Therefore, there are 6 different pairs of snacks that Mr. A may choose.
Answer: (D) 6
When Dima divide the number of digits in the decimal form of 102018 by 4, the remainder is
A) 3 B) 2 C) 1 D) 0
The number of digits in the decimal form of 102018 is 2019; 2019/4 is 504R3.
A) 3
A Matthew sequence is a sequence in which the last number multiply by 5, then minuses by seven. For example: 2, 3, 8, 33, and so on. Elizabeth does not like this sequence, so she makes a new sequence called the Anti-Matthew sequence. In this sequence, you minus 7 the number and then multiply by five. If the Anti-Matthew sequence starts at the number 5, what would be the 5th number in the sequence
A)2033 B)-11710 C)-2335 D)50783 E)10158
5-7*5 = -10, -10-7*5 = -85, -85-7*5 = -460, -460-7*5 = -2335, -2335-7*5 = -11710
Answer is B)-11710
What is the 31st prime number
127
The mean (average) of a set of six numbers is 10. When the number 25 is removed from the set, the mean of the remaining numbers is
(A) 6 (B) 7 (C) 8 (D) 9 (E) 10
The mean (average) of the set of six numbers is 10, and so the sum of the set of six numbers is 6 × 10 = 60. If the number 25 is removed from the set, the sum of the set of the remaining five numbers is 60 − 25 = 35. The mean (average) of the remaining set of five numbers is 35 ÷ 5 = 7.
Answer: (B) 7
Denis bowled on 2 days every week, on a different pair of days each week that Denis bowled. For at most how many weeks did Denis bowl?
A) 14 B) 21 C) 28 D) 35
Each day can be paired with 6 other days for a total of 42 pairs. However, each pair has been counted twice, so there are 21 pairs.
B) 21
Gary just found out that in the numbers 1, 2, 3, 4 the number pattern is +1He also found out that for the numbers 5, 7, 6, 8, 7 the number pattern is +2 then -1.Gary now needs to find the number pattern for: 3, 9, 27, 243, 6561. Can you help Gary find the right answer?
A)Multiplied by 3 B)Multiplied by the last 2 numbers C)The last number squared, cubed, power of 4... D)The last 2 numbers added up and multiplied by 3 E) No pattern
If you look at the first 2 numbers, you can see that when you multiply them it gives you the next number. 3*9=27. To check furthermore, 9*27=243 and 27*243=6561.
Answer is B)Multiplied by the last 2 numbers
What is the first odd number that doesn't have an "e" in it
Trick question, none!
Daniel has a jar that contains 2 red marbles, 2 blue marbles, and no other marbles. He randomly draws 2 marbles from the jar. If the marbles are the same colour, he discards one and puts the other back into the jar. If the marbles are different colours, he discards the red marble and puts the blue marble back into the jar. He repeats this process a total of three times. What is the probability that the remaining marble is red? (BTW the answer is a page long)
(A) 1/2 (B) 1/4 (C) 2/3 (D) 1/3 (E) 0
I WARNED YOU!
Solution 1 Let the letter R represent a red marble, and the letter B represent a blue marble. On his first draw, Daniel may draw RR, RB or BB.
Case 1: Daniel draws RR or RB on his first draw. If Daniel draws RR or RB on his first draw, then he discards the R and the three remaining marbles in the jar are RBB. On his second draw, Daniel may draw RB or BB. If he draws RB, then he discards the R and the two remaining marbles in the jar are BB. Since there are no red marbles remaining, it is not possible for the final marble to be red in this case. If on his second draw Daniel instead draws BB, then he discards a B and the two remaining marbles in the jar are RB. When these are both drawn on her third draw, the R is discarded and the final marble is blue. Again in this case it is not possible for the final marble to be red. Thus, if Daniel draws RR or RB on her first draw, the probability that the final marble is red is zero.
Case 2: Daniel draws BB on his first draw. If Daniel draws BB on his first draw, then he discards a B and the three remaining marbles in the jar are RRB. On his second draw, Daniel may draw RR or RB. If he draws RR or RB, then he discards one R and the two remaining marbles in the jar are RB. When these are both drawn on her third draw, the R is discarded and the final marble is blue. In this case it is not possible for the final marble to be red. Thus, if Daniel draws BB on his first draw, the probability that the final marble is red is zero. Therefore, under the given conditions of drawing and discarding marbles, the probability that Daniel’s last remaining marble is red is zero.
Solution 2 Let the letter R represent a red marble, and the letter B represent a blue marble. If the final remaining marble is R, then the last two marbles must include at least one R. That is, the last two marbles must be RB or RR. If the last two marbles are RB, then when they are drawn from the jar, the R is discarded and the B would remain. Thus it is not possible for the final marble to be R if the final two marbles are RB. So the final remaining marble is R only if the final two marbles are RR. If the final two marbles are RR, then the last three marbles are BRR (since there are only two Rs in the jar at the beginning). However, if the final three marbles are BRR, then when Daniel draws two of these marbles from the jar, at least one of the marbles must be R and therefore one R will be discarded leaving BR as the final two marbles in the jar. That is, it is not possible for the final two marbles in the jar to be RR. The only possibility that the final remaining marble is R occurs when the final two marbles are RR, but this is not possible. Therefore, under the given conditions of drawing and discarding marbles, the probability that Daniel’s last remaining marble is red is zero.
Answer: (E) 0
What is the difference between the product and the sum of the nonzero digits of 2010 when it is written in decimal form? A) 1 B) 2 C) 102 D) 2*10
When expanded, 2010 = 10240 000 000 000. The difference between the product and the sum of the non-zero digits is 8 – 7 = 1.
A) 1
Kati is a very picky eater. In the jar, there are 3 gummy worms, 13 lollipops, 10 russian chewy candy and only 7 chocolate. When she takes candy from her candy jar, she makes sure it's only chocolate. If it is, she eats it. If it isn't, she throws it out without a second thought. If she repeats this process 7 times in the same jar listed above, what is the chance she gets 1 chocolate?
A)71.43% B)15.15% C)68.34% D)14.29% E)3.03%
Probability of getting one chocolate after 7 processes = 1 - (1 - 0.1515)7 = 0.6834 or 68.34%
Answer is C)68.34%
What is the 2nd last prime number before 6500
6481