# Multiple Choice 1 | 7,8,9&10 | CXC CSEC Mathematics | Number Theory

For CXC CSEC Mathematics multiple-choice examinations,
questions 7 to 10 are based on the topic Number Theory. In this video, we will go through the questions
from one past paper. To get the most out of this video, pause it
here, and resume when you have an answer for each question. Give it your best try and if you are stuck
on any, then narrow it down and guess. There is no penalty in this exam for wrong
answers so always choose an answer to each question, even if that means guessing. Last chance to pause the video! I am going to show the working in 5, 4, 3,
2, 1. Okay, let’s go. (7) The LARGEST prime number that is less
than 100 is (A) 91 (B) 93 (C) 97 (D)99. A prime number has only 2 factors: 1 and itself
(see the description below for my videos on prime numbers and factors). My approach to this question is to identify
which numbers are not prime numbers and thus exclude them from the list (of answers). I use the prime numbers, starting with the
smallest to do so. I have explained why I use prime numbers in
a separate video and the link is in the description below. I start with the first prime number, 2. Is 2 a factor of any of these numbers? No, multiples of two are all even numbers
and end in either 0, 2, 4, 6 or 8. None of these numbers is even. The next prime number is 3. If the sum of the individual digits of a number
is a multiple of 3 then the initial number is also a multiple of 3. In 91, 9 + 1=10. 10 is not a multiple of 3, so 91 is not a
multiple of 3. 93: 9 + 3=12. 12 is a multiple of 3, so 93 is also a multiple
of 3. Thus 93 is not a prime number and we can cross
it out. 97: 9 + 7=16. 16 is not a multiple of 3. 99: 9 + 9=18. 18 is a multiple of 3 so we cross 99 out. The answer is either 91 or 97! 5: the multiples
of 5 all end in either 5 or 0. None of these is a multiple of 5. 7 – I do not know a short cut for determining
if a number is a multiple of 7 but let us keep adding 7 to a known multiple of 7 to
generate the other multiples. We know that 70 is a multiple of 7 and if
we keep adding 7 we will get 77, 84, 91, 98… wait!… 91?… 91 is a multiple of 7 and is not a prime number
either. By elimination we have determined that the
answer is (C) 97. (8) The H.C.F. of 12, 15 and 60 is (A) 1 (B)
3 (C) 12 (D) 60. The link to my video on H.C.F. is in the description
below. To determine the H.C.F. we list the factors
and see which factors are common. The factors of 12 are: 1, 2, 3, 4,… we need
to make sure that we do not miss any of the factors – factors exists in pairs. We know that 3 x 4=12 so then the next factor
would be what 2 multiplies by to give 12… 6… and then 1 x 12 gives 12. We now have all the factors. The pairs of factors are 3 and 4, 2 and 6
and 1 and 12. What about the factors of 15? Factors of 15 are: 1, (2 cannot go into 15)
3, (4 cannot go into 15) 5 (we realize that 3 and 5 are a pair of factors so the only
other factor of 15 would be the number that pairs with 1) 1 and 15. The common factors are 1 and 3. 1 is common and 3 is common; no other number
is common. That rules out answers C and D. Both 1 and
3 are factors of 60 so 1 and 3 are the common factors of 12, 15 and 60. The highest one, though, is 3. The answer is B.
(9) By the distributive law 49 x 17 + 49 x 3=(A) 52 + 66 (B) 52 x 66 (C) 49 + 20 (D)
49 x 20. The distributive law says that multiplication
is distributive over addition. i.e. 5 x (3 + 1)=5 x 3 + 5 x 1. In this case, we are given 49 x 17 + 49 x
3. This is equal to 49 x (17 + 3). None of their answers is 49 x (17 + 3) but
if we simplify using BODMAS, working out the brackets, gives us 49 x (17 + 3=) 20. The answer is D.
(10) The value of the digit 5 in the number 537 is (A) 5 (B) 100 (C) 500 (D) 5000
The digit 5 is in the (ones… tens…) hundreds column. The 5 represents 5 hundreds and that is the
answer, C. The link to my video on Place value is in the description below. That is it for this video, my name is Mr.
Dube, see you in the next video.