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18 Jan 2022, 03:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
67% (03:13) correct
33%
(02:54)
wrong
based on 94
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A palindrome is a number that reads the same backward and forward, such as 545. What is the sum of all three-digit palindromes?
A. 49320
B. 49410
C. 49500
D. 49590
E. 49680
Are You Up For the Challenge: 700 Level Questions
A. 49320
B. 49410
C. 49500
D. 49590
E. 49680
Are You Up For the Challenge: 700 Level Questions
21 Jan 2022, 01:12
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I doubt it could be solved in under 3minutes!.. Please let me know if there is a shorter method
for 100s, we have following palindromes... 101 and 111,121, 131.... 191. We can split this as 100+11, 100+21, 100+31.. sum of this is equal to 100*9 + 11+21+...91
Similarly for 2s, we have 202, 212, 222, 232, 242... 292.. We can again split this as 200+11+1, 200+21+2, 200+11+3... This will result in 200 + 11, 200 +21... 200 +91, + 9. Sum of this is 200*9 + 11+21+31...+91 + 9
For 300s, this will result in 300+11, 300+21, 300+21 + 2*9..
Going for other numbers, we will find a similar pattern. So we will have 4 APs altogether...
101+202+303+...+909 ; 900+1800+2700...+8100; 11+21+31+...+91; and 0+9+18+..+72. Note that 11+21+31+..+91 series is repeated 9 times so we need to multiply sum of this series by 9.
Sum of AP = Middle term * no. of terms
Hence,
Sum of
AP 1 = 505*9; AP 2 = 4500*9; AP 3 = 51*9*9; AP 4 = 36*9
= 9 * (505+4500+459+35)
= 9* (5500)
= 495000
(C) IMO
for 100s, we have following palindromes... 101 and 111,121, 131.... 191. We can split this as 100+11, 100+21, 100+31.. sum of this is equal to 100*9 + 11+21+...91
Similarly for 2s, we have 202, 212, 222, 232, 242... 292.. We can again split this as 200+11+1, 200+21+2, 200+11+3... This will result in 200 + 11, 200 +21... 200 +91, + 9. Sum of this is 200*9 + 11+21+31...+91 + 9
For 300s, this will result in 300+11, 300+21, 300+21 + 2*9..
Going for other numbers, we will find a similar pattern. So we will have 4 APs altogether...
101+202+303+...+909 ; 900+1800+2700...+8100; 11+21+31+...+91; and 0+9+18+..+72. Note that 11+21+31+..+91 series is repeated 9 times so we need to multiply sum of this series by 9.
Sum of AP = Middle term * no. of terms
Hence,
Sum of
AP 1 = 505*9; AP 2 = 4500*9; AP 3 = 51*9*9; AP 4 = 36*9
= 9 * (505+4500+459+35)
= 9* (5500)
= 495000
(C) IMO
21 Jan 2022, 02:30
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There will be 9 such series 1a1, 2a2 …. 9a9
For the first one - there will be 10 numbers and the sum can be written as 100*(1+1…. 10 times)+ 10*(0+1+2…9) + (1+1….. 10 times)
For the others, the form will be similar
100x (2+2+2… 10 times) + 10x (0+1+2….9) + (2+2… 10 times)
So the sum can be written as 100*10*(1+2+3….9) + 10*9*(1+2+3…..9) + 10*(1+2…..9)
= 1000*45 + 90*45 + 10*45 = 45000 + 4050 + 450 = 49500 …. (C)
Posted from my mobile device
For the first one - there will be 10 numbers and the sum can be written as 100*(1+1…. 10 times)+ 10*(0+1+2…9) + (1+1….. 10 times)
For the others, the form will be similar
100x (2+2+2… 10 times) + 10x (0+1+2….9) + (2+2… 10 times)
So the sum can be written as 100*10*(1+2+3….9) + 10*9*(1+2+3…..9) + 10*(1+2…..9)
= 1000*45 + 90*45 + 10*45 = 45000 + 4050 + 450 = 49500 …. (C)
Posted from my mobile device
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