<p>What is the least positive integer greater than 7 that leaves a
remainder of 6 when divided by both 9 and 15?</p>
<p>If an integer is divisible by 3, 7, 21, and 49, what is the next
larger integer divisible by these numbers?</p>
<p>Im pretty good at math, but I honestly have no idea how to do these using gcf and lcm concepts.</p>
<p>Good news! You don’t need to know how to compute gcf and lcm - just have your calculator do it for you. For the first one, in your TI-84 calculator press MATH, scroll right to NUM, and press 8 for lcm. Then type 9,15) and you will get an output of 45. So lcm(9,15)=45. So 45 is the least integer divisible by 9 and 15. Now just add 6: 45 + 6 = 51.</p>
<p>For the second one we want lcm(3,7,21,49). Umfortunately your calculator can only do 2 at a time. So as in the last problem, first compute lcm(3,7)=21. Then compute lcm(21,49)=147. So the answer is 147.</p>
<p>For the second one, the calculator only accepts 2 numbers at a time, so how would i go about solving this problem?</p>
<p>In case you’re interested in the actual math, here are two methods for finding the gcd and lcm of 9 and 15:</p>
<p>Method 1: The factors of 9 are 1, 3 and 9. The factors of 15 are 1, 3, 5 and 15. So the common factors of 9 and 15 are 1 and 3. So gcd(9,15) = 3.</p>
<p>The multiples of 9 are 9, 18, 27, 36, 45, 54, 63,… and the multiples of 15 are 15, 30, 45,… We can stop at 45 because 45 is also a multiple of 9. So lcm(9,15) = 45.</p>
<p>Method 2: The prime factorizations of 9 and 15 are 9 = 3^2 and 15 = 3<em>5. To find the gcd we multiply together the smallest powers of each prime from both factorizations, and for the lcm we multiply the highest powers of each prime. So gcd(9,15) = 3 and lcm(9,15) = 3^2</em>5 = 45.</p>
<p>Note: If you have trouble seeing where the gcd and lcm are coming from here, it may help to insert the “missing” primes. In this case, 5 is missing from the factorization of 9. So it might help to write 9 = 3^2<em>5^0. Now we can think of the gcd as 3^1</em>5^0 = 3.</p>
<p>Thanks dr steve. Im still confused about the last problem though</p>
<p>Sent from my SCH-I535 using CC</p>
<p>I’m assuming your confused with the calculator method. It’s unfortunate that the TI-84 can only take the lcm of 2 numbers at a time. I have no idea why the calculator has this limitation - seems like a major oversight on the part of Texas Instruments. But anyway it’s what we’re stuck with until they update their calculator.</p>
<p>Let’s find the lcm of 2, 3, 5, and 14.</p>
<p>We first use our calculator to find the lcm of the first 2 numbers: lcm(2,3)=6.</p>
<p>Now we take the lcm of the result with the 3rd number: lcm(6,5)=30.</p>
<p>Finally, we take the lcm of this new result with the 4th number: lcm(30,14)=210.</p>
<p>You can actually also nest all the lcms together into a single computation:</p>
<p>lcm(lcm(lcm(2,3),5),14)=210.</p>
<p>But I don’t necessarily recommend this nest - it’s likely to lead to human error.</p>
<p>If anyone knows why the TI-84 has this limitation when it comes to computing gcd and lcm, I’d love to hear the reason.</p>
<p>Thank you very much dr steve!</p>
<p>Sent from my SCH-I535 using CC</p>
<p>If an integer is divisible by 3, 7, 21, and 49, what is the next
larger integer divisible by these numbers?</p>
<p>I don’t think this is 147. I would say it is 294. Why? Well, it says an integer is divisible by 3,7,21,49. Well the smallest integer that fits this description is 147. But doesn’t it ask for the next larger integer? Well, then you would get 294.</p>
<p>@Moving</p>
<p>It is just a poorly worded question - clearly not an official College Board question. If your interpretation were correct, then there is no unique solution. </p>
<p>For example, 0 is divisible by 3, 7, 21, and 49. In this case, 147 would be the next larger integer divisible by these numbers.</p>
<p>As another example, -147 is is divisible by 3, 7, 21, and 49. In this case, 0 would be the next larger integer divisible by these numbers.</p>
<p>If that was what they wanted, then the question would have to be worded something like this:</p>
<p>Let k be the least positive integer divisible by 3, 7, 21, and 49. What is the least integer greater than k that is also divisible by these numbers?</p>
<p>I’ve never seen such a thing appear on an SAT.</p>