Digit sums of successive integers
For a natural number $x$ both, the digit sum of $x$ and the digit sum of $x+1$ are multiples of $7$. What is the smallest possible $x$?
mathematics no-computers number-theory
asked yesterday
A. P.A. P.
3,53711145
add a comment |
For a natural number $x$ both, the digit sum of $x$ and the digit sum of $x+1$ are multiples of $7$. What is the smallest possible $x$?
mathematics no-computers number-theory
asked yesterday
A. P.A. P.
3,53711145
"0∉N" That is false.
– Acccumulation
23 hours ago
$0 notin mathbb N$ is true or false depending on the application/author/etc.
– tilper
22 hours ago
Before anyone starts a religious war I better delete this statement. rot13(Vg jnf naljnl bayl zrnag gb thvqr crbcyr vagb gur jebat qverpgvba.)
– A. P.
22 hours ago
add a comment |
For a natural number $x$ both, the digit sum of $x$ and the digit sum of $x+1$ are multiples of $7$. What is the smallest possible $x$?
mathematics no-computers number-theory
asked yesterday
A. P.A. P.
3,53711145
For a natural number $x$ both, the digit sum of $x$ and the digit sum of $x+1$ are multiples of $7$. What is the smallest possible $x$?
mathematics no-computers number-theory
mathematics no-computers number-theory
asked yesterday
A. P.A. P.
3,53711145
asked yesterday
A. P.A. P.
3,53711145
edited 22 hours ago
A. P.
asked yesterday
A. P.A. P.
3,53711145
asked yesterday
A. P.A. P.
3,53711145
asked yesterday
A. P.A. P.
3,53711145
3,53711145
"0∉N" That is false.
– Acccumulation
23 hours ago
$0 notin mathbb N$ is true or false depending on the application/author/etc.
– tilper
22 hours ago
Before anyone starts a religious war I better delete this statement. rot13(Vg jnf naljnl bayl zrnag gb thvqr crbcyr vagb gur jebat qverpgvba.)
– A. P.
22 hours ago
add a comment |
"0∉N" That is false.
– Acccumulation
23 hours ago
$0 notin mathbb N$ is true or false depending on the application/author/etc.
– tilper
22 hours ago
Before anyone starts a religious war I better delete this statement. rot13(Vg jnf naljnl bayl zrnag gb thvqr crbcyr vagb gur jebat qverpgvba.)
– A. P.
22 hours ago
"0∉N" That is false.
– Acccumulation
23 hours ago
"0∉N" That is false.
– Acccumulation
23 hours ago
$0 notin mathbb N$ is true or false depending on the application/author/etc.
– tilper
22 hours ago
$0 notin mathbb N$ is true or false depending on the application/author/etc.
– tilper
22 hours ago
Before anyone starts a religious war I better delete this statement. rot13(Vg jnf naljnl bayl zrnag gb thvqr crbcyr vagb gur jebat qverpgvba.)
– A. P.
22 hours ago
Before anyone starts a religious war I better delete this statement. rot13(Vg jnf naljnl bayl zrnag gb thvqr crbcyr vagb gur jebat qverpgvba.)
– A. P.
22 hours ago
add a comment |
2 Answers
2
active
oldest
votes
69999 (42) and 70000 (7)
...
No two consecutive integers are both multiples of 7, so this needs to take place at a rollover.
...
Rolling over a single 9 drops the digit sum by 8, which isn't a multiple of 7 either.
...
Similarly, if Y=X+1, X99->Y00 drops by 17 and X999->Y000 drops 26.
...
X9999 to Y0000 is the first drop (35) which is itself a multiple of 7...
...
Any number of 9's that's congruent mod 7 to 4 will work, but they'll be much larger, so the first instance must roll over 4 9's.
...
From there, all that remains is to find the first multiple of 10000 with an appropriate digit sum.
...
My initial, less confidence-inspiring method just recognized that 7*10^n was a likely candidate for x+1, and so I started appending 9's to a single 6 until the sum worked out...
answered yesterday
ZomulgustarZomulgustar
1,868723
1
As this is a 'no-computers' puzzle, could you elaborate on how you find this number? Most likely you will also see whether it's minimal if you go through these steps.
– A. P.
yesterday
add a comment |
Suppose x+1 is of the form a*10^n, where the last digit of a is not 0. Then both digitsum(a) and digitsum(a-1)+9*n are multiples of 7. Since digitsum(a-1) = digitsum(a)-1, and the difference between any two multiples of 7 is itself a multiple of 7, we have that 9*n-1 is a multiple of 7. 9*n-1 = (7+2)n-1 = 7n+2n-1, which is a multiple of 7 iff 2n-1 is a multiple of 7. So for the smallest number, we should take 2n-1 = 7, which makes n equal to 4. The smallest possible value of a is 7, giving x = 69999.
answered 23 hours ago
AcccumulationAcccumulation
387110
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
69999 (42) and 70000 (7)
...
No two consecutive integers are both multiples of 7, so this needs to take place at a rollover.
...
Rolling over a single 9 drops the digit sum by 8, which isn't a multiple of 7 either.
...
Similarly, if Y=X+1, X99->Y00 drops by 17 and X999->Y000 drops 26.
...
X9999 to Y0000 is the first drop (35) which is itself a multiple of 7...
...
Any number of 9's that's congruent mod 7 to 4 will work, but they'll be much larger, so the first instance must roll over 4 9's.
...
From there, all that remains is to find the first multiple of 10000 with an appropriate digit sum.
...
My initial, less confidence-inspiring method just recognized that 7*10^n was a likely candidate for x+1, and so I started appending 9's to a single 6 until the sum worked out...
answered yesterday
ZomulgustarZomulgustar
1,868723
1
As this is a 'no-computers' puzzle, could you elaborate on how you find this number? Most likely you will also see whether it's minimal if you go through these steps.
– A. P.
yesterday
add a comment |
69999 (42) and 70000 (7)
...
No two consecutive integers are both multiples of 7, so this needs to take place at a rollover.
...
Rolling over a single 9 drops the digit sum by 8, which isn't a multiple of 7 either.
...
Similarly, if Y=X+1, X99->Y00 drops by 17 and X999->Y000 drops 26.
...
X9999 to Y0000 is the first drop (35) which is itself a multiple of 7...
...
Any number of 9's that's congruent mod 7 to 4 will work, but they'll be much larger, so the first instance must roll over 4 9's.
...
From there, all that remains is to find the first multiple of 10000 with an appropriate digit sum.
...
My initial, less confidence-inspiring method just recognized that 7*10^n was a likely candidate for x+1, and so I started appending 9's to a single 6 until the sum worked out...
answered yesterday
ZomulgustarZomulgustar
1,868723
1
As this is a 'no-computers' puzzle, could you elaborate on how you find this number? Most likely you will also see whether it's minimal if you go through these steps.
– A. P.
yesterday
add a comment |
69999 (42) and 70000 (7)
...
No two consecutive integers are both multiples of 7, so this needs to take place at a rollover.
...
Rolling over a single 9 drops the digit sum by 8, which isn't a multiple of 7 either.
...
Similarly, if Y=X+1, X99->Y00 drops by 17 and X999->Y000 drops 26.
...
X9999 to Y0000 is the first drop (35) which is itself a multiple of 7...
...
Any number of 9's that's congruent mod 7 to 4 will work, but they'll be much larger, so the first instance must roll over 4 9's.
...
From there, all that remains is to find the first multiple of 10000 with an appropriate digit sum.
...
My initial, less confidence-inspiring method just recognized that 7*10^n was a likely candidate for x+1, and so I started appending 9's to a single 6 until the sum worked out...
answered yesterday
ZomulgustarZomulgustar
1,868723
69999 (42) and 70000 (7)
...
No two consecutive integers are both multiples of 7, so this needs to take place at a rollover.
...
Rolling over a single 9 drops the digit sum by 8, which isn't a multiple of 7 either.
...
Similarly, if Y=X+1, X99->Y00 drops by 17 and X999->Y000 drops 26.
...
X9999 to Y0000 is the first drop (35) which is itself a multiple of 7...
...
Any number of 9's that's congruent mod 7 to 4 will work, but they'll be much larger, so the first instance must roll over 4 9's.
...
From there, all that remains is to find the first multiple of 10000 with an appropriate digit sum.
...
My initial, less confidence-inspiring method just recognized that 7*10^n was a likely candidate for x+1, and so I started appending 9's to a single 6 until the sum worked out...
answered yesterday
ZomulgustarZomulgustar
1,868723
edited yesterday
answered yesterday
ZomulgustarZomulgustar
1,868723
answered yesterday
ZomulgustarZomulgustar
1,868723
answered yesterday
ZomulgustarZomulgustar
1,868723
1,868723
1
As this is a 'no-computers' puzzle, could you elaborate on how you find this number? Most likely you will also see whether it's minimal if you go through these steps.
– A. P.
yesterday
add a comment |
1
As this is a 'no-computers' puzzle, could you elaborate on how you find this number? Most likely you will also see whether it's minimal if you go through these steps.
– A. P.
yesterday
1
1
As this is a 'no-computers' puzzle, could you elaborate on how you find this number? Most likely you will also see whether it's minimal if you go through these steps.
– A. P.
yesterday
As this is a 'no-computers' puzzle, could you elaborate on how you find this number? Most likely you will also see whether it's minimal if you go through these steps.
– A. P.
yesterday
add a comment |
Suppose x+1 is of the form a*10^n, where the last digit of a is not 0. Then both digitsum(a) and digitsum(a-1)+9*n are multiples of 7. Since digitsum(a-1) = digitsum(a)-1, and the difference between any two multiples of 7 is itself a multiple of 7, we have that 9*n-1 is a multiple of 7. 9*n-1 = (7+2)n-1 = 7n+2n-1, which is a multiple of 7 iff 2n-1 is a multiple of 7. So for the smallest number, we should take 2n-1 = 7, which makes n equal to 4. The smallest possible value of a is 7, giving x = 69999.
answered 23 hours ago
AcccumulationAcccumulation
387110
add a comment |
Suppose x+1 is of the form a*10^n, where the last digit of a is not 0. Then both digitsum(a) and digitsum(a-1)+9*n are multiples of 7. Since digitsum(a-1) = digitsum(a)-1, and the difference between any two multiples of 7 is itself a multiple of 7, we have that 9*n-1 is a multiple of 7. 9*n-1 = (7+2)n-1 = 7n+2n-1, which is a multiple of 7 iff 2n-1 is a multiple of 7. So for the smallest number, we should take 2n-1 = 7, which makes n equal to 4. The smallest possible value of a is 7, giving x = 69999.
answered 23 hours ago
AcccumulationAcccumulation
387110
add a comment |
Suppose x+1 is of the form a*10^n, where the last digit of a is not 0. Then both digitsum(a) and digitsum(a-1)+9*n are multiples of 7. Since digitsum(a-1) = digitsum(a)-1, and the difference between any two multiples of 7 is itself a multiple of 7, we have that 9*n-1 is a multiple of 7. 9*n-1 = (7+2)n-1 = 7n+2n-1, which is a multiple of 7 iff 2n-1 is a multiple of 7. So for the smallest number, we should take 2n-1 = 7, which makes n equal to 4. The smallest possible value of a is 7, giving x = 69999.
answered 23 hours ago
AcccumulationAcccumulation
387110
Suppose x+1 is of the form a*10^n, where the last digit of a is not 0. Then both digitsum(a) and digitsum(a-1)+9*n are multiples of 7. Since digitsum(a-1) = digitsum(a)-1, and the difference between any two multiples of 7 is itself a multiple of 7, we have that 9*n-1 is a multiple of 7. 9*n-1 = (7+2)n-1 = 7n+2n-1, which is a multiple of 7 iff 2n-1 is a multiple of 7. So for the smallest number, we should take 2n-1 = 7, which makes n equal to 4. The smallest possible value of a is 7, giving x = 69999.
answered 23 hours ago
AcccumulationAcccumulation
387110
answered 23 hours ago
AcccumulationAcccumulation
387110
answered 23 hours ago
AcccumulationAcccumulation
387110
answered 23 hours ago
AcccumulationAcccumulation
387110
387110
add a comment |
add a comment |
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"0∉N" That is false.
– Acccumulation
23 hours ago
$0 notin mathbb N$ is true or false depending on the application/author/etc.
– tilper
22 hours ago
Before anyone starts a religious war I better delete this statement. rot13(Vg jnf naljnl bayl zrnag gb thvqr crbcyr vagb gur jebat qverpgvba.)
– A. P.
22 hours ago