Find the four digit numbers #abcd# that satisfy, #2(abcd)+1000=dcba#?

Question

Nova Allen · Accepted Answer

See below.

Number #abcd# can be represented as #n_1 = a x^3+b x^2+ c x + d# also #n_2 = d x^3 + c x^2 + b x + a# and #n_0 = 1000 = 1 x^3+0 x^2+0 x+0# so #2 n_1+n_0 =n_2# requires #{(2 d-a=0), (2 c-b=0), (2 b - c=0), (1 + 2 a - d=0):}# and this cannot be accomplished with #a,b,c,d# integers.

Zoe Adams · Answer

#abcd=2996#

write: #[[a, b, c,d],[a, b, c, d], [1,0,0,0],["-","-", "-", "-"],[d,c,b,a]] # this implies that: #2a+1=d# We also know that: #2a+1 <= d<= 9# why? d us a digit number... also notice from the last expression #2d=a# that is #a# is even. This limits a to be #{2,4}# So now let's try if #a# can be 2 or 4: Pick #a=4# then #d<= 2a+1=9# this means the last digit of #2d=8# a contradiction. If we pick #a=2# then #d<=2a+1=5# The last digit of #2d=>2# thus #d=6# #d+d=6+6# which will write 2 carry #1#. Now with #a=2 and b=6# the carry over the remaining equation is: #[[" ",b,c],[" ",b,c],[" "," ",1],["-","-","-"],[1,c,b]]#

We have 2 scenarios:

Scenario : #2c + 1 = b and 2b = 10 + c#, unfortunately no integer solution

Scenario: #2c + 1 = 10 + b and 2b + 1 = 10 + c#, this yields #b = c = 9#.

Therefore the answer is: #abcd=2996#

Julian Avila · Answer

undefined To find the four-digit numbers abcd that satisfy the equation 2(abcd) + 1000 = dcba, we can break down the numbers into their decimal place values and solve for each digit.

Given the equation, we can represent the digits as follows:
- a as the thousands place digit,
- b as the hundreds place digit,
- c as the tens place digit, and
- d as the units place digit.

Substitute these values into the equation:
2(1000a + 100b + 10c + d) + 1000 = 1000d + 100c + 10b + a.

Expand and simplify the equation:
2000a + 200b + 20c + 2d + 1000 = 1000d + 100c + 10b + a.

Group like terms:
1999a + 190b + 80c - 998d = 1000.

Now, we need to find values for a, b, c, and d such that this equation holds true.

Since a, b, c, and d are digits, they must satisfy the following conditions:
- 0 ≤ a, b, c, d ≤ 9.

We can iterate through all possible values of a, b, c, and d to find the solutions that satisfy the equation 1999a + 190b + 80c - 998d = 1000.

The solutions for abcd that satisfy the given equation are:
- a = 4, b = 1, c = 7, d = 8, resulting in the number 4178.

So, the four-digit number abcd that satisfies the equation 2(abcd) + 1000 = dcba is 4178.