Abou the last exercise in which you affirm that x*y > 6, I disagree because the number 4 is a counterexample, since 4 divided by 5 is 0 and the remainder is 4. So there is a number that is less than 6 and still the remainder of the integer division by 5 is 4.
when he says that xy>6, he means that xy has to be above 6, but that doesn't mean it is ALL the numbers above 6, only the ones that apply. For example, he showed that if xy = 19, the remainder is 4, so it works. At the same time, if xy = 7, like you say, then the remainder is 2, so it doesn't work. The question asks you to assume that the numbers in question all lead to remainder 4 when divided by 5.
i have a question. i have a question here that reads "when 15 is divided by the positive integer k, the remainder is 3. for how man different values of k is the this true?" i was wondering if you had any tips on how to solve this without trying to list all the possible numbers..? thanks for your help
Jonathan Hashmonay I remember this question from the SAT Blue Book and I don't think there's a better way to do the problem other than quickly running through the numbers up to 15.
+Jonathan Hashmonay 15-3 is 12, so any factors of 12 will work, 12,6,4,3,2,1. However, the remainder is 3, so 1,2,3 will never have a remainder of 3. so 12,6,4 are the answers.
QINNIYQ Interesting thing, you actually don't have to add when the variable is on top--you can subtract in both scenarios. If you chose to subtract in the second problem, it would look like this: k - 2 / 3 = 2 k = 8 and 8 divided by 3 still gives a valid remainder of 2 2k = 16 and 16 divided by 3 also gives a remainder of 1 Not sure why he chose to add :T
Great tutorials, keep up the good work! There's one little thing that I don't get though: What if you would divide 4 by 5? (xy = 4) Wouldn't that end up with 0 rem 4?
For the last one, xy could equal 4. 5 goes into 4 zero times with a remainder of 4. If you wanted to fix this problem you could say that x and y must be unique numbers.
Here is one I found Step 1: Perform the division in your calculator: 14/4 = 3.5 Step 2: Multiply the integer part of this answer by the divisor: 4*3 = 12 Step 3: Subtract the above result from the dividend to get the remainder: 14 - 12 = 2. Thus the remainder is 2.
Hey guys think about the first question. The value of x can be 6 as well. Six times two equals 12 leaving a remainder of 4. So the value of x+1 can also be 7.
I still don't understand the last one. Like, it says which of these must be true? But aren't they all true? Kind of a stupid question for me to ask though. Should know by now. :/
What if, for the last problem, xy = 4? 5 divides evenly into zero, which is 4 less than 4. I believe you got that wrong. The correct answer was F) None of the above.
Last question : 4 divided by 5 gives remainder of 4. So xy > 6 CANNOT be the answer. According to me, Option A is the correct answer because 9/5 definitely gives a remainder of 4.
man, i thought remainders were the toughest shit to figure out!!! but they're SOOO EASY now that you've explained it!
I will tell all my friends to follow you , you deserve it !!
Ahmed Abdulla thanks a lot, hope you did well on your exam
thanks for your help, this really boost my morale, I thought these questions were like messing up my rhythm in the SATS.
i'll just watch a bunch of your videos like a day before the test so i can remember your examples and tips
Abou the last exercise in which you affirm that x*y > 6, I disagree because the number 4 is a counterexample, since 4 divided by 5 is 0 and the remainder is 4. So there is a number that is less than 6 and still the remainder of the integer division by 5 is 4.
For the last question i was puzzled!... The answer you said was xy>6... But what if xy is 7(which is greater than 6) should that be false as well?
when he says that xy>6, he means that xy has to be above 6, but that doesn't mean it is ALL the numbers above 6, only the ones that apply. For example, he showed that if xy = 19, the remainder is 4, so it works. At the same time, if xy = 7, like you say, then the remainder is 2, so it doesn't work. The question asks you to assume that the numbers in question all lead to remainder 4 when divided by 5.
should I check or is it always correct
how did it go?
man you make it look so easy
i have a question. i have a question here that reads
"when 15 is divided by the positive integer k, the remainder is 3. for how man different values of k is the this true?"
i was wondering if you had any tips on how to solve this without trying to list all the possible numbers..?
thanks for your help
Jonathan Hashmonay I remember this question from the SAT Blue Book and I don't think there's a better way to do the problem other than quickly running through the numbers up to 15.
+Jonathan Hashmonay
15-3 is 12, so any factors of 12 will work, 12,6,4,3,2,1. However, the remainder is 3, so 1,2,3 will never have a remainder of 3. so 12,6,4 are the answers.
also, on the last one, can't you just read the options and identify the last one is impossible right away?
At the last why E? 7 and 8 does not work.If I started at E, I could easily cross out E and go for D. Reply would be grate.
Can number one also equal 7?
What about the fact that you subtracted when the variable was on the bottom and then you added when the variable was on top.
QINNIYQ Interesting thing, you actually don't have to add when the variable is on top--you can subtract in both scenarios. If you chose to subtract in the second problem, it would look like this:
k - 2 / 3 = 2
k = 8
and 8 divided by 3 still gives a valid remainder of 2
2k = 16
and 16 divided by 3 also gives a remainder of 1
Not sure why he chose to add :T
Great tutorials, keep up the good work!
There's one little thing that I don't get though:
What if you would divide 4 by 5? (xy = 4)
Wouldn't that end up with 0 rem 4?
I totally agree
The answer to this question is no solution. None of these statements are true.
for the K divided by 3 problem.
I got 14 divided by 3 equals remainder 2
2(14)=28
28 divided by 3 equals remainder 1
Is it still right?
For the last one, xy could equal 4. 5 goes into 4 zero times with a remainder of 4. If you wanted to fix this problem you could say that x and y must be unique numbers.
salsadancer00 it still wouldn't work; 4 = 4x1 .: x & y are unique
Samuel Hošovský you're right. This problem's screwed!
Here is one I found
Step 1: Perform the division in your calculator: 14/4 = 3.5
Step 2: Multiply the integer part of this answer by the divisor: 4*3 = 12
Step 3: Subtract the above result from the dividend to get the remainder:
14 - 12 = 2.
Thus the remainder is 2.
got a 510 need a 700 in two days. THANK YOU SIR
if 16 is divided by 6 then the remainder is still 4... can this one be correct???
This helped a lot!!Thank you c:
for the last question, the correct answer is a rather than e in my opinion
I think in the last question (e) option i.e xy>6 is also wrong RadicalPrep
Thank you so much!
I understand it now, but once I take the test I'll probably forget these techniques :P
1 is the remainder when 10 ism divided by 3. 9/3 is 3 and then 1 is left over
I always get these wrong, thanks!
Thank you so much!! :)
5/3 is not giving me a remainder of 2, it's giving me 1.6 please explain, thank you D:
+Rosemary Ochoa lol. remainder only is given when you do LONG division. he said in the video to not use the calculator. it wont give you the remainder
+Frank Lopez OMG i feel dumb. I was stressed out when xD thank you
+Rosemary Ochoa no problem. good luck on the test, im taking mine tomorrow lol
+Frank Lopez me too, I don't feel prepared at all, but lets do it. hope you do great too
Hey guys think about the first question. The value of x can be 6 as well. Six times two equals 12 leaving a remainder of 4. So the value of x+1 can also be 7.
Aditya Ram that is correct. There are infinite possible answers for that question but it asked which number is one of the possibilities.
Samuel Hošovský sorry but i think x can take only two values 6 and 12 to satisfy the above condition not infinite values... cheers
I still don't understand the last one. Like, it says which of these must be true? But aren't they all true? Kind of a stupid question for me to ask though. Should know by now. :/
thanks
The last one was probably the easiest :)
the last one in my opinion was easy 6 divided by 5 can never have a remainder of 4, so it had be be higher.
the last example was actually wrong. xy can equal 4. 4 divided by 5 yields a remainder of 4.
Cant XY = 4? X=4 Y=1
XY/5 = 0 remainder 4
What if, for the last problem, xy = 4? 5 divides evenly into zero, which is 4 less than 4. I believe you got that wrong. The correct answer was F) None of the above.
Last question : 4 divided by 5 gives remainder of 4.
So xy > 6 CANNOT be the answer.
According to me, Option A is the correct answer because 9/5 definitely gives a remainder of 4.
Oh wow. I just realized why I didn't get it. It's cuz' it HAS to be it. Wow.
thx alooot
the first problem i divided 16 by 6 and got a remainder of 4 and got x+1=7
EXAM TOMORROWWWWWWWW YIKESSSSSSSSSSSSSS
I was doing this question with my brother who had a 600 on the Math Portion on the SATs and this got him angry . Lol
SAT tomorrow damn im scared
I am lost with number 1
Omg you rule
i understand it. its not that hard.
these questions are WAYYYYYY to easy for me.
You know why? Because i'm an Indian and India is in Asia.
(no squinty eyes though)
can you be a teacher at my school please we'll pay you with money and food
no lol because 5 doesn't fit in 4, you have no remainder