By Enzo - Harrow Hong Kong
Firstly, to make 747 by adding a 3 digit number and its reversal, there are 3 different ways:
126 + 621
225 + 522
324 + 423
And the rest are repeated.
Now the first digit often equals the last digit, and the middle digit must be even because you are adding the same tens digit in this case (10b+10b=20b). We hence evaluate out 707, 727, 747, 767, and 787. The tens digit is 0, 1, 2, 3, and 4 respectively, and the hundreds digit does not change. Therefore, these are the easiest to find out since there is no carrying involved.
However, in a different case, the hundreds digit could possibly be carried from the tens digit. This happens when the tens digit is equal to or greater than 5 since 5+5=10. Hence the units digit will be 7-1=6 in these conditions.
Consequently, we have the following possible solutions: 706, 726, 746, 766, and 786. Here, the tens digits of the original numbers are both 5, 6, 7, 8, and 9 respectively. The 7s in the hundreds place all come from a 6 that was carried by the tens place.
Right now, I am going to show you a few possible ways that it can be done.
707=106+601
727=116+611
747=126+621
767=136+631
787=146+641
706=155+551
726=165+561
746=175+571
766=185+581
786=195+591
I understand there are a lot more methods to get these results, but here are some examples.
For 300 to 400, it is the same scenario, except that the numbers start with a 1 or 2. We compute the following numbers:
303=102+201
323=112+211
343=122+221
363=132+231
383=142+241
302=151+151
322=161=161
342=171+171
362=181+181
382=191+191
And the same is true for those from 800 to 900:
808=107+701
828=117+711
848=127+721
868=137+731
888=127+741
807=156+651
827=166+661
847=176+671
867=186+681
887=196+691
We now move on to the four-digit numbers.
To make 1251, take a 3-digit number and write it as 100a+10b+c. Its reversal is 100c+10b+a. Adding them up gives us 101a+20b+101c, which is also 101(a+c)+20b. The main problem is the 20b, hence we are going to subtract 20 each time from our desired four-digit number until we get one that is divisible by 101.
In this case, 1251 is not divisible by 101, so we subtract 20 from the number each time until we get to a number that can be divided by 101.
We have 1231, 1211, 1191, 1171, 1151, 1131, 1111. We stop at 1111 because it is divisible by 101 (101*11). We have subtracted in 20s for 7 times (1251-7*20=1111). Hence, the tens digit must be 7, and the units and hundreds digits must add up to 11. Here we have 4 different cases that it could be done:
279+972=1251
378+873=1251
477+774=1251
576+675=1251
From the totals from 1200 to 1300, each number can be calculated using 101(a+c)+20b, where c is an integer from 0 to 9. Since we have 1111 as the base number (101 * 11), there include 1211, 1231, 1251, 1271, and 1291, where b is equal to 5, 6, 7, 8, and 9. It is also possible to have 101 * 12 = 1212, hence we include 1212, 1232, 1252, 1272, and 1292 for this to be possible. In this case, b is equal to those from 0 to 4.
Here are a few methods to achieve these results.
1211=259+952
1231=269+962
1251=279+972
1271=289+982
1291=299+992
1212=309+903
1232=319+913
1252=329+923
1272=339+933
1292=349+943
Totals from 1900 to 2000: Since a and b have to be single-digit integers, the largest possible value for 101(a+c) = 101*18 = 1818. Here, b will need to be from 5 to 9 in order to make the total greater than 1900 but less than 2000, or else, the first two digits will remain as “18”. This time, only five numbers satisfy the condition.
1918=959+959
1938=969+969
1958=979+979
1978=989+989
1998=999+999
If there were ten totals, there would be a case that a+c=19. However, the greatest value that a+c can take is 18, because a and c are both equal to 9. This is true because both a and c have to be single-digit integers (largest is 9).
>>> POSSIBLE EXTENSION
We now consider subtracting a number from its reversal. Firstly, we continue to use the three-digit numbers 100a+10b+c and 100c+10b+a. Subtracting them gives us 99a-99c. This is much easier than adding because there are less options to consider. For those between 200 and 300, there is only one number than is divisible by 99, and that is 297.
603-306=297
From 1200 to 1300, we have four digit numbers 1000a+100b+10c+d and 1000d+100c+10b+a. Subtracting gives us 999a+90b-90c-999d. From this specific range, we first concentrate on 999a-999d. The first result is when a-d=1, which gives 999. We add 90 every time until we get a four-digit integer from 1200 to 1300. This number, in turn, is 999+(3*90)=1269.
An example to show this is as follows: 2521-1252=1269. There are obviously more ways to get 1269, but here is one of them shown above.
We move on to a second case, which is when a-d=2, which gives 1998. This time, we subtract 90 every time because there is no restriction on whether we can add or subtract (unlike the 3-digit numbers) as the middle two digits can be different.
Subtracting 90 until we get a number within the range (1200, 1300):
1998-90=1908
1908-90=1818
1818-90=1728
1728-90=1638
1638-90=1548
1548-90=1458
1458-90=1368
1368-90=1278.
We stop at 1278 because it is within the desired range. In this case, b is less than c, and they differ by exactly 8. Therefore b=1 and c=9. Here is a way to show this.
3191-1913=1278
Did you know? There are exactly 161 unique four-digit numbers that can be obtained by subtracting a four digit number by its reversal, ranging from 1089 to 8991.