Tìm tất cả các số chính phương có tổng các chữ số của nó bằng 2013?
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Bài này có trong tạp chí Toán Tuổi Thơ
Bài này của lớp 6
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địt lồn mẹ mày bú buồi đầu buồi liếm lông chim
a) 102; 111; 120; 210; 201; 300
b) 4000 3100 3010 3001 2002 2020 2200 2011 2101 2110 1111 1003 1030 1300 1210 1201 1120 1102 1012 1021 v.v..
Nếu thiểu bổ sung hộ mk.
Gọi số đó là abc ta có abc=a+b+c. Giả sử \(9\ge a\ge b\ge c\ge0\) thì \(abc=a+b+c\le3a\) suy ra \(bc\le3\)
abc khác 0 (vì \(a\ge1\) nên a+b+c\(\ge1\))
bc=3 thì b=3, c=1, suy ra 3+1+a=3a suy ra a=2 (loại, vì a bé hơn b)
bc=2 => b=2, c=1 suy ra 3+a=2a suy ra a=3 (Chọn)
bc=1 => b=1, c=1 suy ra 2+a=a (loại)
Vậy Các số cần tìm là 123, 132, 213, 231, 312, 321
Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath
đúng ko
gọi số cần tìm là ab,(a khác 0;a,b<10)
ta có:
ab+ba=10a+b+10b+aq=11a+11b=11(a+b)
vì a+b là số chính phương nên a+b chia hết cho 11
mà 1<a<10
0<b<10
=> 1<a+b<20
=>a+b=11
ta có bảng sau:
a | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
b | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 |
vây có 8 số thỏa mãn đề bài