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The threedigit positive integer x has the hundreds,
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15 Nov 2013, 08:00
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The threedigit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The threedigit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y? A) 21 B) 200 C) 210 D) 300 E) 310
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The threedigit positive integer x has the hundreds,
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15 Nov 2013, 08:09
mohnish104 wrote: The threedigit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The threedigit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?
A) 21 B) 200 C) 210 D) 300 E) 310 x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively. y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively. \((2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)\); \(2^{ak}*3^{bl}*5^{cm}=2^2*3\); \(ak=2\), \(bl=1\), and \(cm=0\); \(xy=210\). Answer: C.
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The threedigit positive integer x has the hundreds,
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02 Dec 2014, 02:00
x = 100a + 10b + c ........... (1) y = 100k + 10l + m ............ (2) We require to find xy (1)  (2) \(2^a* 3^b * 5^c = 12 * 2^k* 3^l * 5^m\) \(2^a * 3^b * 5^c = 2^{k+2} * 3^{l+1} * 5^m\) Equating powers a = k+2; b = l+1; c = m Substituting the above values in equation (1) 100k + 200 + 10l + 10 + c  (100k + 10l + m) = 210 Answer = C
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Re: The threedigit positive integer x has the hundreds,
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15 Nov 2013, 08:13
Bunuel wrote: mohnish104 wrote: The threedigit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The threedigit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?
A) 21 B) 200 C) 210 D) 300 E) 310 x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively. y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively. \((2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)\) > \(2^{ak}*3^{bl}*5^{cm}=2^2*3\) > \(ak=2\), \(bl=1\), and \(cm=0\) > \(xy=210\). Answer: C. Similar questions to practice: kandlareeachfourdigitpositiveintegerswiththousands126646.htmlthefunctionfisdefinedforeachpositivethreedigit100847.htmlforanyfourdigitnumberabcdabcd3a5b7c11d126522.htmlHope this helps.
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Re: The threedigit positive integer x has the hundreds,
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09 Dec 2014, 08:58
Bunuel wrote: mohnish104 wrote: The threedigit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The threedigit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?
A) 21 B) 200 C) 210 D) 300 E) 310 x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively. y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively. \((2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)\) > \(2^{ak}*3^{bl}*5^{cm}=2^2*3\) > \(ak=2\), \(bl=1\), and \(cm=0\) > \(xy=210\). Answer: C. I am sorry but i dont understand the slightest thing of what's going on here. How does that suddenly translate to 2^2*3? really confused here.



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Re: The threedigit positive integer x has the hundreds,
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09 Dec 2014, 09:03
JoostGrijsen wrote: Bunuel wrote: mohnish104 wrote: The threedigit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The threedigit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?
A) 21 B) 200 C) 210 D) 300 E) 310 x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively. y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively. \((2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)\) > \(2^{ak}*3^{bl}*5^{cm}=2^2*3\) > \(ak=2\), \(bl=1\), and \(cm=0\) > \(xy=210\). Answer: C. I am sorry but i dont understand the slightest thing of what's going on here. How does that suddenly translate to 2^2*3? really confused here. \((2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)\); \(\frac{(2^a)(3^b)(5^c)}{(2^k)(3^l)(5^m)} = 12\); \(2^{ak}*3^{bl}*5^{cm}=2^2*3\). Similar questions to practice: kandlareeachfourdigitpositiveintegerswiththousands91004.htmlforanyfourdigitnumberabcdabcd3a5b7c11d126522.htmlkandlareeachfourdigitpositiveintegerswiththousands126646.htmlthefunctionfisdefinedforeachpositivethreedigit100847.htmlforathreedigitnumberxyzwherexyandzarethe59284.htmlHope it helps.
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Re: The threedigit positive integer x has the hundreds,
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31 Mar 2016, 18:11
Attached is a visual that should help.
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Screen Shot 20160331 at 6.10.08 PM.png [ 137.01 KiB  Viewed 8437 times ]
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Re: The threedigit positive integer x has the hundreds,
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28 Apr 2017, 04:57
what is the next step after this 2a−k∗3b−l∗5c−m=22∗32a−k∗3b−l∗5c−m=22∗3 ? I cant figure it out. can someone break this down well for me please



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Re: The threedigit positive integer x has the hundreds,
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11 Aug 2017, 02:48
(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m) can be re written as (2^ak)(3^bl)(5^cm)=2^2.3^1.5^0 ak=2 bl=1 cm=0 The above 3 are derived from laws of indices, which in turn points to the subtraction of each of the unit's,ten's and hundred's place of x & y. Option C.



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Re: The threedigit positive integer x has the hundreds,
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15 Aug 2017, 17:57
Bunuel wrote: mohnish104 wrote: The threedigit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The threedigit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?
A) 21 B) 200 C) 210 D) 300 E) 310 x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively. y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively. \((2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)\) > \(2^{ak}*3^{bl}*5^{cm}=2^2*3\) > \(ak=2\), \(bl=1\), and \(cm=0\) > \(xy=210\). Answer: C. please clarify my doubt: does the wording sound good? The threedigit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. it would be better if it was asam i right? previous statement can be interpreted as there is a positive integer a, whose hundreds digit is the hundred digit of x. please clarify!!



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Re: The threedigit positive integer x has the hundreds,
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24 Aug 2017, 08:10
Can someone please explain me after this step 2a−k∗3b−l∗5c−m



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The threedigit positive integer x has the hundreds,
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27 Aug 2018, 06:38
Bunuel wrote: mohnish104 wrote: The threedigit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The threedigit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?
A) 21 B) 200 C) 210 D) 300 E) 310 x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively. y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively. \((2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)\) > \(2^{ak}*3^{bl}*5^{cm}= 2^2*3\) > \(ak=2\), \(bl=1\), and \(cm=0\) > \(xy=210\).Answer: C. ok i got it how you got\(2^2*3\) but i still dont get how you got this part \(ak=2\), \(bl=1\), and \(cm=0\) > \(xy=210\) Hey pushpitkc are you there ?



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The threedigit positive integer x has the hundreds,
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27 Aug 2018, 12:33
dave13 wrote: Bunuel wrote: mohnish104 wrote: The threedigit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The threedigit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?
A) 21 B) 200 C) 210 D) 300 E) 310 x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively. y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively. \((2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)\) > \(2^{ak}*3^{bl}*5^{cm}= 2^2*3\) > \(ak=2\), \(bl=1\), and \(cm=0\) > \(xy=210\).Answer: C. ok i got it how you got\(2^2*3\) but i still dont get how you got this part \(ak=2\), \(bl=1\), and \(cm=0\) > \(xy=210\) Hey pushpitkc are you there ? Hey dave13I've always been here We have \((2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)\) and can be rewritten as \(\frac{(2^a)(3^b)(5^c)}{(2^k)(3^l)(5^m)} = 12\) > (1) Now, since we know that \(\frac{2^a}{2^k} = 2^{ak}\)  \(\frac{3^b}{3^l} = 3^{bl}\)  \(\frac{5^c}{5^m} = 5^{cm}\). Also, 12 can be rewritten as \(2^2*3^1*5^0\) Equation (1) now becomes > a  k = 2  b  l = 1  c  m = 0 and since we need the difference of 3 digit numbers abc(100a + 10b + c) and klm(100k + 10l + m) > 2*100 + 1*10 + 0 = 210 Hope this clears your confusion!
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Re: The threedigit positive integer x has the hundreds,
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28 Aug 2018, 01:24
I don't get why bl = 1 etc, can someone please break this part down



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The threedigit positive integer x has the hundreds,
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28 Aug 2018, 02:21
onyx12102 wrote: I don't get why bl = 1 etc, can someone please break this part down Hey onyx12102Please look at this part of my previous solution LHS => \(\frac{2^a}{2^k} + \frac{3^b}{3^l} + \frac{5^c}{5^m}\) = \(2^{ ak} +3^{ bl} + 5^{ cm}\). Also, RHS which is 12 can be rewritten as \(2^ 2*3^ 1*5^ 0\) Equating equal bases on both sides, we will get a  k = 3 , b  l = 1 , and c  m = 0 Hope this clears your confusion!
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The threedigit positive integer x has the hundreds,
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