Two 3 digit numbers are formed using the same integers and when they are added they form another 3 digit number with the same integers, find the added number?
Answers
Answer:
I am going to discuss three numbers 1089, 1729 (Ramanujan Number),6174 (Kaprekar's Number)
Magic Number 1089
In base 10, the following steps always yield 1089:
Take any three-digit number where the first and last digits differ by 2 or more.
Reverse the digits, and subtract the smaller from the larger one.
Add to this result the number produced by reversing its digits.
For example, if the spectator chooses 237 (or 732):
732 − 237 = 495
495 + 594 = 1089
Explanation
If we let a, b, c denote the three digits of the original number, then the three-digit number is 100a+10b+c. The reverse is 100c+10b+a. Subtract: (100a+10b+c)-(100c+10b+a) to get 99(a-c). Since the digits were decreasing, (a-c) is at least 2 and no greater than 9, so the result must be one of 198, 297, 396, 495, 594, 693, 792, or 891. When you add any one of those numbers to the reverse of itself, you get 1089
Other properties
Multiplying the number 1089 by the integers from 1 to 9 produces a pattern: multipliers adding up to 10 give products that are the digit reversals of each other:
1 × 1089 = 1089 ↔ 9 × 1089 = 9801
2 × 1089 = 2178 ↔ 8 × 1089 = 8712
3 × 1089 = 3267 ↔ 7 × 1089 = 7623
4 × 1089 = 4356 ↔ 6 × 1089 = 6534
5 × 1089 = 5445 ↔ 5 × 1089 = 5445
Also note the patterns within each column:
1 × 1089 = 1089
2 × 1089 = 2178
3 × 1089 = 3267
4 × 1089 = 4356
5 × 1089 = 5445
6 × 1089 = 6534
7 × 1089 = 7623
8 × 1089 = 8712
9 × 1089 = 9801
Numbers formed analogously in other bases, e.g. octal 1067 or hexadecimal 10EF, also have these properties.
Number 1729:
1729 is the Hardy–Ramanujan number after a famous anecdote of the British mathematician.
It is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways.
1729=13+123=93+103
1729 is also the third Carmichael number and the first absolute Euler pseudoprime. It is also a sphenic number.
1729 is a Zeisel number. It is a centered cube number, as well as a dodecagonal number, a 24-gonal and 84-gonal number.
Investigating pairs of distinct integer-valued quadratic forms that represent every integer the same number of times, Schiemann found that such quadratic forms must be in four or more variables, and the least possible discriminant of a four-variable pair is 1729 (Guy 2004).
Because in base 10 the number 1729 is divisible by the sum of its digits, it is a Harshad number. It also has this property in octal ( 1729=33018, 3 + 3 +0+1=7 ) and hexadecimal ( 1729=6C116,6+C+1=1910 ), but not in binary.
In base 12, 1729 is written as 1001, so its reciprocal has only period 6 in that base.
1729 has another mildly interesting property: the 1729th decimal place is the beginning of the first consecutive occurrence of all ten digits without repetition in the decimal representation of the transcendental number e.
Masahiko Fujiwara showed that 1729 is one of four positive integers (with the others being 81, 1458, and the trivial case 1) which, when its digits are added together, produces a sum which, when multiplied by its reversal, yields the original number:
1 + 7 + 2 + 9 = 19
19 × 91 = 1729
It suffices only to check sums congruent to 0 or 1 (mod 9) up to 19.
Number 6174:
6174 is known as Kaprekar's constant.
after the Indian mathematician D. R. Kaprekar. This number is notable for the following property:
Take any four-digit number, using at least two different digits. (Leading zeros are allowed.)
Arrange the digits in descending and then in ascending order to get two four-digit numbers, adding leading zeros if necessary.
Subtract the smaller number from the bigger number.
Go back to step 2.
The above process, known as Kaprekar's routine, will always reach its fixed point, 6174, in at most 7 iterations.
Once 6174 is reached, the process will continue yielding 7641 – 1467 = 6174. For example, choose 3524:
5432 – 2345 = 3087
8730 – 0378 = 8352
8532 – 2358 = 6174
7641 – 1467 = 6174
The only four-digit numbers for which Kaprekar's routine does not reach 6174 are repdigits such as 1111, which give the result 0000after a single iteration. All other four-digit numbers eventually reach 6174 if leading zeros are used to keep the number of digits at 4:
2111 – 1112 = 0999
9990 – 0999 = 8991 (rather than 999 – 999 = 0)
9981 – 1899 = 8082
8820 – 0288 = 8532
8532 – 2358 = 6174
9831 reaches 6174 after 7 iterations:
9831 – 1389 = 8442
8442 – 2448 = 5994
9954 – 4599 = 5355
5553 – 3555 = 1998
9981 – 1899 = 8082
8820 – 0288 = 8532 (rather than 882 – 288 = 594)
8532 – 2358 = 6174
4371 reaches 6174 after 7 iterations:
7431 - 1347 = 6084
8640 - 0468 = 8172 (rather than 864 - 468 = 396)
8721 - 1278 = 7443
7443 - 3447 = 3996
9963 - 3699 = 6264
6642 - 2466 = 4176
7641 - 1467 = 6174
8774, 8477, 8747, 7748, 7487, 7847, 7784, 4877, 4787, and 4778 reach 6174 after 4 iterations:
8774 – 4778 = 3996
9963 – 3699 = 6264
6642 – 2466 = 4176
7641 – 1467 = 6174
Answer:
890
Step-by-step explanation: